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Exterior derivative of differential p-form $\omega$ can be defined by "(p+1)-linear part of the value of $\omega$ integrated over the boundary of infinitesimal (p+1)-parallelotope".

More specifically, $$d\omega(v_1,v_2,...,v_{p+1})\\ =\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}\omega$$

where $[tv_1,tv_2,...,tv_{p+1}]$ is (p+1)-parallelotope spanned by $tv_1,tv_2,...,tv_{p+1}$.

This aspect of exterior derivative is already mentioned by Petya and MathCrawler, but there is no proof why it is equal to the standard definition of exterior derivative . So I'll give you.

It suffices to prove that

$$d\big(f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\big)(v_1,v_2,...,v_{p+1})\\ =\sum_{i\in \{ 1,2,3,...,n \} }\frac{f(x_1,x_2,...,x_n)}{\partial x_i} x_i\wedge x_1\wedge x_2\wedge ...\wedge x_p (v_1,v_2,...,v_{p+1})$$

is equal to $$\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$$

Suppose $U\subset \mathbb{R^n} $ is open, $\sigma^t:[0,t]^{p+1}\to U$ is $C^{\infty}$and, $\mathbf{t} \mapsto \big(\sigma^t_1(\mathbf{t}),\sigma^t_2(\mathbf{t}),\sigma^t_3(\mathbf{t}),...,\sigma^t_n(\mathbf{t})\big)\in U$ $f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$ is differential p-form on $U$

In order to define $\partial\sigma^t$ with induced orientation, let $$d^j_-(t_1, \dots, t_{p+1})=(t_1, \dots, t_{j-1}, 0, t_{j+1}, \dots, t_{p+1}),\\ d^j_+(t_1, \dots, t_{p-1}) = (t_1, \dots, t_{j-1}, t, t_{j+1}, \dots, t_{p+1}).$$

Then $\partial \sigma^t = \sum_{j=1}^{p+1} (-1)^j(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)$, and it can be computed as below

$$ \begin{align} &\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma^t}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^{p+1}(-1)^j\int_{(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^{p+1}(-1)^j\Big(\int_{\sigma^t\circ d^j_-} f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p \\ &-\int_{\sigma^t\circ d^j_+} f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\Big)\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^p(-1)^j\Big(\int_{[0,t]^p}f(\sigma^t\circ d^j_-)det \begin{vmatrix} \frac{\sigma^t_1\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_-}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_-}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}\circ d^j_-}{\partial t_1} &\cdots & \frac{\sigma^t_l\circ d^j_-}{\partial t_{j-1}}& \frac{\sigma^t_l\circ d^j_-}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l\circ d^j_-}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{p+1}} \end{vmatrix}dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p -\int_{[0,t]^p}f(\sigma^t\circ d^j_+)det \begin{vmatrix} \frac{\sigma^t_1\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_+}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_+}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}\circ d^j_+}{\partial t_1} &\cdots & \frac{\sigma^t_l\circ d^j_+}{\partial t_{j-1}}& \frac{\sigma^t_l\circ d^j_+}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_k\circ d^j_+}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{p+1}} \end{vmatrix} dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\Big)\\ \Big(=&\bigstar\Big) \end{align} $$

To Continue the computation, I will use a following fact.

$$\begin{align} &\lim_{t\to 0}\frac1{t^{p+1}}\int_{[0,t]^p}g(d^t_+)-g(d^t_-) dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\\ =&\lim_{t\to 0}\frac1{t^{p+1}}\int_{[0,t]^p}g(t_1,\dots ,t_{j-1},t,t_{j+1},\dots,t_{p+1})-g(t_1,\dots ,t_{j-1},0,t_{j+1},\dots,t_{p+1}) dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\\ =&\frac{\partial}{\partial t_j}g(t_1,\dots ,t_j,\dots,t_{p+1})\rvert_{t_{i=0}}\quad \Big(=\frac{\partial g}{\partial t_j}(0,\dots,0)\Big) \end{align}$$

This fact can be proved by iterated integration and taylor expansion around $(0,\dots ,0)$.

Then, $$\begin{align} \Big(\bigstar\Big) =&\sum_{j=1}^p(-1)^{j+1}\frac{\partial}{\partial t_j}f(\sigma^t)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{j=1}^{p+1}(-1)^{j+1}\Big( \sum_{k=1}^n\frac{\partial f}{\partial x_k}\frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ +&\sum_{j=1}^{p+1}(-1)^{j+1}f(\sigma^t)\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{k=1}^n\frac{\partial f}{\partial x_k}\sum_{j=1}^{p+1}(-1)^{j-1}\Big( \frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ +&f(\sigma^t)\sum_{j=1}^{p+1}(-1)^{j-1}\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =\big(&\bigstar\bigstar\big) \end{align}$$

To continue the calculation further more, I will use two formula.

1. $$\begin{align}&\sum_{j=1}^{p+1}(-1)^{j-1}\Big( \frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&det \begin{vmatrix} \frac{\sigma^t_k}{\partial t_1} & \cdots & \frac{\sigma^t_k}{\partial t_{j-1}} &\frac{\sigma^t_k}{\partial t_{j}}&\frac{\sigma^t_k}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_k}{\partial t_{p+1}}\\ \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} & \frac{\sigma^t_1}{\partial t_{j}}&\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}} & \frac{\sigma^t_l}{\partial t_{j}} & \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} & \frac{\sigma^t_{p}}{\partial t_{j}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&dx_k\wedge dx_1\wedge dx_2\wedge\dots\wedge dx_p (\frac{\sigma^t}{\partial t_{1}},\frac{\sigma^t}{\partial t_{2}},\dots,\frac{\sigma^t}{\partial t_{p+1}})_{\Big\lvert_{t_i=0}} \end{align}$$

2. $$\begin{align} &\sum_{j=1}^{p+1}(-1)^{j-1}\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{\tau\in S_{p+1}} \big(sgn(\tau)\big) \frac{\sigma^t_{1}}{\partial t_{\tau(1)}\partial t_{\tau(2)}} \frac{\sigma^t_{2}}{\partial t_{\tau(3)}} \frac{\sigma^t_{3}}{\partial t_{\tau(4)}}\dots \frac{\sigma^t_{p}}{\partial t_{\tau(p+1)}} _{\Big\lvert_{t_i=0}}\\ =&0 \\ &\big( \text{since}\quad sgn(\tau)\frac{\sigma^t_{1}}{\partial t_{\tau(1)}\partial t_{\tau(2)}} =-sgn(\tau{'})\frac{\sigma^t_{1}}{\partial t_{\tau{'}(1)}\partial t_{\tau{'}(2)}} \quad \text{where}\quad \tau(1)=\tau^{'}(2),\tau(l)=\tau^{'}(l)\text{for} \quad l\geq3\big) \end{align}$$

Applying 1.and 2.to$\big(\bigstar\bigstar\big)$,I finally get $$ \begin{align} &\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma^t}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\sum_{k=1}^n\frac{\partial f}{\partial x_k} dx_k\wedge dx_1\wedge dx_2\wedge\dots\wedge dx_p (\frac{\sigma^t}{\partial t_{1}},\frac{\sigma^t}{\partial t_{2}},\dots,\frac{\sigma^t}{\partial t_{p+1}})_{\Big\lvert_{t_i=0}} \end{align} $$

Then when $\sigma^t(t_1,t_2,\dots ,t_{p+1})=\sum_i t_iv_i\in U$, and $\sigma^t(0)=(x_1,x_2,\dots,x_n)$,this is what I needed first.

Exterior derivative of differential p-form $\omega$ can be defined by "(p+1)-linear part of the value of $\omega$ integrated over the boundary of infinitesimal (p+1)-parallelotope".

More specifically, $$d\omega(v_1,v_2,...,v_{p+1})\\ =\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}\omega$$

where $[tv_1,tv_2,...,tv_{p+1}]$ is (p+1)-parallelotope spanned by $tv_1,tv_2,...,tv_{p+1}$.

This aspect of exterior derivative is already mentioned by Petya and MathCrawler, but there is no proof why it is equal to the standard definition of exterior derivative . So I'll give you.

It suffices to prove that

$$d\big(f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\big)(v_1,v_2,...,v_{p+1})\\ =\sum_{i\in \{ 1,2,3,...,n \} }\frac{f(x_1,x_2,...,x_n)}{\partial x_i} x_i\wedge x_1\wedge x_2\wedge ...\wedge x_p (v_1,v_2,...,v_{p+1})$$

is equal to $$\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$$

Suppose $U\subset \mathbb{R^n} $ is open, $\sigma^t:[0,t]^{p+1}\to U$ is $C^{\infty}$and, $\mathbf{t} \mapsto \big(\sigma^t_1(\mathbf{t}),\sigma^t_2(\mathbf{t}),\sigma^t_3(\mathbf{t}),...,\sigma^t_n(\mathbf{t})\big)\in U$ $f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$ is differential p-form on $U$

In order to define $\partial\sigma^t$ with induced orientation, let $$d^j_-(t_1, \dots, t_{p+1})=(t_1, \dots, t_{j-1}, 0, t_{j+1}, \dots, t_{p+1}),\\ d^j_+(t_1, \dots, t_{p-1}) = (t_1, \dots, t_{j-1}, t, t_{j+1}, \dots, t_{p+1}).$$

Then $\partial \sigma^t = \sum_{j=1}^{p+1} (-1)^j(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)$, and it can be computed as below

$$ \begin{align} &\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma^t}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^{p+1}(-1)^j\int_{(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^{p+1}(-1)^j\Big(\int_{\sigma^t\circ d^j_-} f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p \\ &-\int_{\sigma^t\circ d^j_+} f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\Big)\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^p(-1)^j\Big(\int_{[0,t]^p}f(\sigma^t\circ d^j_-)det \begin{vmatrix} \frac{\sigma^t_1\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_-}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_-}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}\circ d^j_-}{\partial t_1} &\cdots & \frac{\sigma^t_l\circ d^j_-}{\partial t_{j-1}}& \frac{\sigma^t_l\circ d^j_-}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l\circ d^j_-}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{p+1}} \end{vmatrix}dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p \\ &-\int_{[0,t]^p}f(\sigma^t\circ d^j_+)det \begin{vmatrix} \frac{\sigma^t_1\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_+}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_+}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}\circ d^j_+}{\partial t_1} &\cdots & \frac{\sigma^t_l\circ d^j_+}{\partial t_{j-1}}& \frac{\sigma^t_l\circ d^j_+}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_k\circ d^j_+}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{p+1}} \end{vmatrix} dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\Big)\\ \Big(=&\bigstar\Big) \end{align} $$

To Continue the computation, I will use a following fact.

$$\begin{align} &\lim_{t\to 0}\frac1{t^{p+1}}\int_{[0,t]^p}g(d^t_+)-g(d^t_-) dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\\ =&\lim_{t\to 0}\frac1{t^{p+1}}\int_{[0,t]^p}g(t_1,\dots ,t_{j-1},t,t_{j+1},\dots,t_{p+1})-g(t_1,\dots ,t_{j-1},0,t_{j+1},\dots,t_{p+1}) dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\\ =&\frac{\partial}{\partial t_j}g(t_1,\dots ,t_j,\dots,t_{p+1})\rvert_{t_{i=0}}\quad \Big(=\frac{\partial g}{\partial t_j}(0,\dots,0)\Big) \end{align}$$

This fact can be proved by iterated integration and taylor expansion around $(0,\dots ,0)$.

Then, $$\begin{align} \Big(\bigstar\Big) =&\sum_{j=1}^p(-1)^{j+1}\frac{\partial}{\partial t_j}f(\sigma^t)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{j=1}^{p+1}(-1)^{j+1}\Big( \sum_{k=1}^n\frac{\partial f}{\partial x_k}\frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ +&\sum_{j=1}^{p+1}(-1)^{j+1}f(\sigma^t)\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{k=1}^n\frac{\partial f}{\partial x_k}\sum_{j=1}^{p+1}(-1)^{j-1}\Big( \frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ +&f(\sigma^t)\sum_{j=1}^{p+1}(-1)^{j-1}\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =\big(&\bigstar\bigstar\big) \end{align}$$

To continue the calculation further more, I will use two formula.

1. $$\begin{align}&\sum_{j=1}^{p+1}(-1)^{j-1}\Big( \frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&det \begin{vmatrix} \frac{\sigma^t_k}{\partial t_1} & \cdots & \frac{\sigma^t_k}{\partial t_{j-1}} &\frac{\sigma^t_k}{\partial t_{j}}&\frac{\sigma^t_k}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_k}{\partial t_{p+1}}\\ \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} & \frac{\sigma^t_1}{\partial t_{j}}&\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}} & \frac{\sigma^t_l}{\partial t_{j}} & \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} & \frac{\sigma^t_{p}}{\partial t_{j}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&dx_k\wedge dx_1\wedge dx_2\wedge\dots\wedge dx_p (\frac{\sigma^t}{\partial t_{1}},\frac{\sigma^t}{\partial t_{2}},\dots,\frac{\sigma^t}{\partial t_{p+1}})_{\Big\lvert_{t_i=0}} \end{align}$$

2. $$\begin{align} &\sum_{j=1}^{p+1}(-1)^{j-1}\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{\tau\in S_{p+1}} \big(sgn(\tau)\big) \frac{\sigma^t_{1}}{\partial t_{\tau(1)}\partial t_{\tau(2)}} \frac{\sigma^t_{2}}{\partial t_{\tau(3)}} \frac{\sigma^t_{3}}{\partial t_{\tau(4)}}\dots \frac{\sigma^t_{p}}{\partial t_{\tau(p+1)}} _{\Big\lvert_{t_i=0}}\\ =&0 \\ &\big( \text{since}\quad sgn(\tau)\frac{\sigma^t_{1}}{\partial t_{\tau(1)}\partial t_{\tau(2)}} =-sgn(\tau{'})\frac{\sigma^t_{1}}{\partial t_{\tau{'}(1)}\partial t_{\tau{'}(2)}} \quad \text{where}\quad \tau(1)=\tau^{'}(2),\tau(l)=\tau^{'}(l)\text{for} \quad l\geq3\big) \end{align}$$

Applying 1.and 2.to$\big(\bigstar\bigstar\big)$,I finally get $$ \begin{align} &\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma^t}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\sum_{k=1}^n\frac{\partial f}{\partial x_k} dx_k\wedge dx_1\wedge dx_2\wedge\dots\wedge dx_p (\frac{\sigma^t}{\partial t_{1}},\frac{\sigma^t}{\partial t_{2}},\dots,\frac{\sigma^t}{\partial t_{p+1}})_{\Big\lvert_{t_i=0}} \end{align} $$

Then when $\sigma^t(t_1,t_2,\dots ,t_{p+1})=\sum_i t_iv_i\in U$, and $\sigma^t(0)=(x_1,x_2,\dots,x_n)$,this is what I needed first.

Exterior derivative of differential p-form $\omega$ can be defined by "(p+1)-linear part of the value of $\omega$ integrated over the boundary of infinitesimal (p+1)-parallelotope".

More specifically, $$d\omega(v_1,v_2,...,v_{p+1})\\ =\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}\omega$$

where $[tv_1,tv_2,...,tv_{p+1}]$ is (p+1)-parallelotope spanned by $tv_1,tv_2,...,tv_{p+1}$.

This aspect of exterior derivative is already mentioned by Petya and MathCrawler, but there is no proof why it is equal to the standard definition of exterior derivative . So I'll give you.

It suffices to prove that

$$d\big(f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\big)(v_1,v_2,...,v_{p+1})\\= \sum_{i\in \{ 1,2,3,...,n \} }\frac{f(x_1,x_2,...,x_n)}{\partial x_i} x_i\wedge x_1\wedge x_2\wedge ...\wedge x_p (v_1,v_2,...,v_{p+1})$$

is equal to $$\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$$

Suppose $U\subset \mathbb{R^n} $ is open, $\sigma^t:[0,t]^{p+1}\to U$ is $C^{\infty}$and, $\mathbf{t} \mapsto \big(\sigma^t_1(\mathbf{t}),\sigma^t_2(\mathbf{t}),\sigma^t_3(\mathbf{t}),...,\sigma^t_n(\mathbf{t})\big)\in U$ $f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$ is differential p-form on $U$

In order to define $\partial\sigma^t$ with induced orientation, let $d^j_-(t_1, \dots, t_{p+1})=(t_1, \dots, t_{j-1}, 0, t_{j+1}, \dots, t_{p+1}),\\ d^j_+(t_1, \dots, t_{p-1}) = (t_1, \dots, t_{j-1}, t, t_{j+1}, \dots, t_{p+1}).$

Then $\partial \sigma^t = \sum_{j=1}^{p+1} (-1)^j(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)$, and it can be computed as below

$$ \begin{align} &\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma^t}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^{p+1}(-1)^j\int_{(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^{p+1}(-1)^j\Big(\int_{\sigma^t\circ d^j_-} f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p -\int_{\sigma^t\circ d^j_+} f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\Big)\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^p(-1)^j\Big(\int_{[0,t]^p}f(\sigma^t\circ d^j_-)det \begin{vmatrix} \frac{\sigma^t_1\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_-}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_-}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}\circ d^j_-}{\partial t_1} &\cdots & \frac{\sigma^t_l\circ d^j_-}{\partial t_{j-1}}& \frac{\sigma^t_l\circ d^j_-}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l\circ d^j_-}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{p+1}} \end{vmatrix}dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p -\int_{[0,t]^p}f(\sigma^t\circ d^j_+)det \begin{vmatrix} \frac{\sigma^t_1\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_+}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_+}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}\circ d^j_+}{\partial t_1} &\cdots & \frac{\sigma^t_l\circ d^j_+}{\partial t_{j-1}}& \frac{\sigma^t_l\circ d^j_+}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_k\circ d^j_+}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{p+1}} \end{vmatrix} dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\Big)\\ \Big(=&\bigstar\Big) \end{align} $$

To Continue the computation, I will use a following fact.

$$\begin{align} &\lim_{t\to 0}\frac1{t^{p+1}}\int_{[0,t]^p}g(d^t_+)-g(d^t_-) dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\\ =&\lim_{t\to 0}\frac1{t^{p+1}}\int_{[0,t]^p}g(t_1,\dots ,t_{j-1},t,t_{j+1},\dots,t_{p+1})-g(t_1,\dots ,t_{j-1},0,t_{j+1},\dots,t_{p+1}) dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\\ =&\frac{\partial}{\partial t_j}g(t_1,\dots ,t_j,\dots,t_{p+1})\rvert_{t_{i=0}}\quad \Big(=\frac{\partial g}{\partial t_j}(0,\dots,0)\Big) \end{align}$$

This fact can be proved by iterated integration and taylor expansion around $(0,\dots ,0)$.

Then, $$\begin{align} \Big(\bigstar\Big) =&\sum_{j=1}^p(-1)^{j+1}\frac{\partial}{\partial t_j}f(\sigma^t)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{j=1}^{p+1}(-1)^{j+1}\Big( \sum_{k=1}^n\frac{\partial f}{\partial x_k}\frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ +&\sum_{j=1}^{p+1}(-1)^{j+1}f(\sigma^t)\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{k=1}^n\frac{\partial f}{\partial x_k}\sum_{j=1}^{p+1}(-1)^{j-1}\Big( \frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ +&f(\sigma^t)\sum_{j=1}^{p+1}(-1)^{j-1}\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =\big(&\bigstar\bigstar\big) \end{align}$$

To continue the calculation further more, I will use two formula.

1. $$\begin{align}&\sum_{j=1}^{p+1}(-1)^{j-1}\Big( \frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&det \begin{vmatrix} \frac{\sigma^t_k}{\partial t_1} & \cdots & \frac{\sigma^t_k}{\partial t_{j-1}} &\frac{\sigma^t_k}{\partial t_{j}}&\frac{\sigma^t_k}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_k}{\partial t_{p+1}}\\ \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} & \frac{\sigma^t_1}{\partial t_{j}}&\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}} & \frac{\sigma^t_l}{\partial t_{j}} & \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} & \frac{\sigma^t_{p}}{\partial t_{j}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&dx_k\wedge dx_1\wedge dx_2\wedge\dots\wedge dx_p (\frac{\sigma^t}{\partial t_{1}},\frac{\sigma^t}{\partial t_{2}},\dots,\frac{\sigma^t}{\partial t_{p+1}})_{\Big\lvert_{t_i=0}} \end{align}$$

2. $$\begin{align} &\sum_{j=1}^{p+1}(-1)^{j-1}\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{\tau\in S_{p+1}} \big(sgn(\tau)\big) \frac{\sigma^t_{1}}{\partial t_{\tau(1)}\partial t_{\tau(2)}} \frac{\sigma^t_{2}}{\partial t_{\tau(3)}} \frac{\sigma^t_{3}}{\partial t_{\tau(4)}}\dots \frac{\sigma^t_{p}}{\partial t_{\tau(p+1)}} _{\Big\lvert_{t_i=0}}\\ =&0 \\ &\big( \text{since}\quad sgn(\tau)\frac{\sigma^t_{1}}{\partial t_{\tau(1)}\partial t_{\tau(2)}} =-sgn(\tau{'})\frac{\sigma^t_{1}}{\partial t_{\tau{'}(1)}\partial t_{\tau{'}(2)}} \quad \text{where}\quad \tau(1)=\tau^{'}(2),\tau(l)=\tau^{'}(l)\text{for} \quad l\geq3\big) \end{align}$$

Applying 1.and 2.to$\big(\bigstar\bigstar\big)$,I finally get $$ \begin{align} &\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma^t}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\sum_{k=1}^n\frac{\partial f}{\partial x_k} dx_k\wedge dx_1\wedge dx_2\wedge\dots\wedge dx_p (\frac{\sigma^t}{\partial t_{1}},\frac{\sigma^t}{\partial t_{2}},\dots,\frac{\sigma^t}{\partial t_{p+1}})_{\Big\lvert_{t_i=0}} \end{align} $$

Then when $\sigma^t(t_1,t_2,\dots ,t_{p+1})=\sum_i t_iv_i\in U$, and $\sigma^t(0)=(x_1,x_2,\dots,x_n)$,this is what I needed first.

Exterior derivative of differential p-form $\omega$ can be defined by "(p+1)-linear part of the value of $\omega$ integrated over the boundary of infinitesimal (p+1)-parallelotope".

More specifically, $$d\omega(v_1,v_2,...,v_{p+1})\\ =\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}\omega$$

where $[tv_1,tv_2,...,tv_{p+1}]$ is (p+1)-parallelotope spanned by $tv_1,tv_2,...,tv_{p+1}$.

This aspect of exterior derivative is already mentioned by Petya and MathCrawler, but there is no proof why it is equal to the standard definition of exterior derivative . So I'll give you.

It suffices to prove that

$$d\big(f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\big)(v_1,v_2,...,v_{p+1})\\ =\sum_{i\in \{ 1,2,3,...,n \} }\frac{f(x_1,x_2,...,x_n)}{\partial x_i} x_i\wedge x_1\wedge x_2\wedge ...\wedge x_p (v_1,v_2,...,v_{p+1})$$

is equal to $$\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$$

Suppose $U\subset \mathbb{R^n} $ is open, $\sigma^t:[0,t]^{p+1}\to U$ is $C^{\infty}$and, $\mathbf{t} \mapsto \big(\sigma^t_1(\mathbf{t}),\sigma^t_2(\mathbf{t}),\sigma^t_3(\mathbf{t}),...,\sigma^t_n(\mathbf{t})\big)\in U$ $f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$ is differential p-form on $U$

In order to define $\partial\sigma^t$ with induced orientation, let $$d^j_-(t_1, \dots, t_{p+1})=(t_1, \dots, t_{j-1}, 0, t_{j+1}, \dots, t_{p+1}),\\ d^j_+(t_1, \dots, t_{p-1}) = (t_1, \dots, t_{j-1}, t, t_{j+1}, \dots, t_{p+1}).$$

Then $\partial \sigma^t = \sum_{j=1}^{p+1} (-1)^j(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)$, and it can be computed as below

$$ \begin{align} &\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma^t}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^{p+1}(-1)^j\int_{(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^{p+1}(-1)^j\Big(\int_{\sigma^t\circ d^j_-} f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p \\ &-\int_{\sigma^t\circ d^j_+} f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\Big)\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^p(-1)^j\Big(\int_{[0,t]^p}f(\sigma^t\circ d^j_-)det \begin{vmatrix} \frac{\sigma^t_1\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_-}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_-}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}\circ d^j_-}{\partial t_1} &\cdots & \frac{\sigma^t_l\circ d^j_-}{\partial t_{j-1}}& \frac{\sigma^t_l\circ d^j_-}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l\circ d^j_-}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{p+1}} \end{vmatrix}dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p -\int_{[0,t]^p}f(\sigma^t\circ d^j_+)det \begin{vmatrix} \frac{\sigma^t_1\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_+}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_+}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}\circ d^j_+}{\partial t_1} &\cdots & \frac{\sigma^t_l\circ d^j_+}{\partial t_{j-1}}& \frac{\sigma^t_l\circ d^j_+}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_k\circ d^j_+}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{p+1}} \end{vmatrix} dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\Big)\\ \Big(=&\bigstar\Big) \end{align} $$

To Continue the computation, I will use a following fact.

$$\begin{align} &\lim_{t\to 0}\frac1{t^{p+1}}\int_{[0,t]^p}g(d^t_+)-g(d^t_-) dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\\ =&\lim_{t\to 0}\frac1{t^{p+1}}\int_{[0,t]^p}g(t_1,\dots ,t_{j-1},t,t_{j+1},\dots,t_{p+1})-g(t_1,\dots ,t_{j-1},0,t_{j+1},\dots,t_{p+1}) dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\\ =&\frac{\partial}{\partial t_j}g(t_1,\dots ,t_j,\dots,t_{p+1})\rvert_{t_{i=0}}\quad \Big(=\frac{\partial g}{\partial t_j}(0,\dots,0)\Big) \end{align}$$

This fact can be proved by iterated integration and taylor expansion around $(0,\dots ,0)$.

Then, $$\begin{align} \Big(\bigstar\Big) =&\sum_{j=1}^p(-1)^{j+1}\frac{\partial}{\partial t_j}f(\sigma^t)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{j=1}^{p+1}(-1)^{j+1}\Big( \sum_{k=1}^n\frac{\partial f}{\partial x_k}\frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ +&\sum_{j=1}^{p+1}(-1)^{j+1}f(\sigma^t)\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{k=1}^n\frac{\partial f}{\partial x_k}\sum_{j=1}^{p+1}(-1)^{j-1}\Big( \frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ +&f(\sigma^t)\sum_{j=1}^{p+1}(-1)^{j-1}\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =\big(&\bigstar\bigstar\big) \end{align}$$

To continue the calculation further more, I will use two formula.

1. $$\begin{align}&\sum_{j=1}^{p+1}(-1)^{j-1}\Big( \frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&det \begin{vmatrix} \frac{\sigma^t_k}{\partial t_1} & \cdots & \frac{\sigma^t_k}{\partial t_{j-1}} &\frac{\sigma^t_k}{\partial t_{j}}&\frac{\sigma^t_k}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_k}{\partial t_{p+1}}\\ \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} & \frac{\sigma^t_1}{\partial t_{j}}&\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}} & \frac{\sigma^t_l}{\partial t_{j}} & \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} & \frac{\sigma^t_{p}}{\partial t_{j}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&dx_k\wedge dx_1\wedge dx_2\wedge\dots\wedge dx_p (\frac{\sigma^t}{\partial t_{1}},\frac{\sigma^t}{\partial t_{2}},\dots,\frac{\sigma^t}{\partial t_{p+1}})_{\Big\lvert_{t_i=0}} \end{align}$$

2. $$\begin{align} &\sum_{j=1}^{p+1}(-1)^{j-1}\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{\tau\in S_{p+1}} \big(sgn(\tau)\big) \frac{\sigma^t_{1}}{\partial t_{\tau(1)}\partial t_{\tau(2)}} \frac{\sigma^t_{2}}{\partial t_{\tau(3)}} \frac{\sigma^t_{3}}{\partial t_{\tau(4)}}\dots \frac{\sigma^t_{p}}{\partial t_{\tau(p+1)}} _{\Big\lvert_{t_i=0}}\\ =&0 \\ &\big( \text{since}\quad sgn(\tau)\frac{\sigma^t_{1}}{\partial t_{\tau(1)}\partial t_{\tau(2)}} =-sgn(\tau{'})\frac{\sigma^t_{1}}{\partial t_{\tau{'}(1)}\partial t_{\tau{'}(2)}} \quad \text{where}\quad \tau(1)=\tau^{'}(2),\tau(l)=\tau^{'}(l)\text{for} \quad l\geq3\big) \end{align}$$

Applying 1.and 2.to$\big(\bigstar\bigstar\big)$,I finally get $$ \begin{align} &\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma^t}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\sum_{k=1}^n\frac{\partial f}{\partial x_k} dx_k\wedge dx_1\wedge dx_2\wedge\dots\wedge dx_p (\frac{\sigma^t}{\partial t_{1}},\frac{\sigma^t}{\partial t_{2}},\dots,\frac{\sigma^t}{\partial t_{p+1}})_{\Big\lvert_{t_i=0}} \end{align} $$

Then when $\sigma^t(t_1,t_2,\dots ,t_{p+1})=\sum_i t_iv_i\in U$, and $\sigma^t(0)=(x_1,x_2,\dots,x_n)$,this is what I needed first.

Exterior derivative of differential p-form $\omega$ can be defined by "(p+1)-linear part of the value of $\omega$ integrated over the boundary of infinitesimal (p+1)-parallelotope".

More specifically, $d\omega(v_1,v_2,...,v_{p+1})=\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}\omega$, where $[tv_1,tv_2,...,tv_{p+1}]$ is (p+1)-parallelotope spanned by $tv_1,tv_2,...,tv_{p+1}$.

This aspect of exterior derivative is already mentioned by Petya and MathCrawler, but there is no proof why it is equal to the standard definition of exterior derivative . So I'll give you.

It suffices to prove that

$$d\big(f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\big)(v_1,v_2,...,v_{p+1})\\= \sum_{i\in \{ 1,2,3,...,n \} }\frac{f(x_1,x_2,...,x_n)}{\partial x_i} x_i\wedge x_1\wedge x_2\wedge ...\wedge x_p (v_1,v_2,...,v_{p+1})$$

is equal to $$\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$$

Suppose $U\subset \mathbb{R^n} $ is open, $\sigma^t:[0,t]^{p+1}\to U$ is $C^{\infty}$and, $\mathbf{t} \mapsto \big(\sigma^t_1(\mathbf{t}),\sigma^t_2(\mathbf{t}),\sigma^t_3(\mathbf{t}),...,\sigma^t_n(\mathbf{t})\big)\in U$ $f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$ is differential p-form on $U$

In order to define $\partial\sigma^t$ with induced orientation, let $d^j_-(t_1, \dots, t_{p+1})=(t_1, \dots, t_{j-1}, 0, t_{j+1}, \dots, t_{p+1}),\\ d^j_+(t_1, \dots, t_{p-1}) = (t_1, \dots, t_{j-1}, t, t_{j+1}, \dots, t_{p+1}).$

Then $\partial \sigma^t = \sum_{j=1}^{p+1} (-1)^j(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)$, and it can be computed as below

$$ \begin{align} &\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma^t}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^{p+1}(-1)^j\int_{(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^{p+1}(-1)^j\Big(\int_{\sigma^t\circ d^j_-} f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p -\int_{\sigma^t\circ d^j_+} f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\Big)\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^p(-1)^j\Big(\int_{[0,t]^p}f(\sigma^t\circ d^j_-)det \begin{vmatrix} \frac{\sigma^t_1\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_-}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_-}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}\circ d^j_-}{\partial t_1} &\cdots & \frac{\sigma^t_l\circ d^j_-}{\partial t_{j-1}}& \frac{\sigma^t_l\circ d^j_-}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l\circ d^j_-}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{p+1}} \end{vmatrix}dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p -\int_{[0,t]^p}f(\sigma^t\circ d^j_+)det \begin{vmatrix} \frac{\sigma^t_1\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_+}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_+}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}\circ d^j_+}{\partial t_1} &\cdots & \frac{\sigma^t_l\circ d^j_+}{\partial t_{j-1}}& \frac{\sigma^t_l\circ d^j_+}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_k\circ d^j_+}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{p+1}} \end{vmatrix} dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\Big)\\ \Big(=&\bigstar\Big) \end{align} $$

To Continue the computation, I will use a following fact.

$$\begin{align} &\lim_{t\to 0}\frac1{t^{p+1}}\int_{[0,t]^p}g(d^t_+)-g(d^t_-) dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\\ =&\lim_{t\to 0}\frac1{t^{p+1}}\int_{[0,t]^p}g(t_1,\dots ,t_{j-1},t,t_{j+1},\dots,t_{p+1})-g(t_1,\dots ,t_{j-1},0,t_{j+1},\dots,t_{p+1}) dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\\ =&\frac{\partial}{\partial t_j}g(t_1,\dots ,t_j,\dots,t_{p+1})\rvert_{t_{i=0}}\quad \Big(=\frac{\partial g}{\partial t_j}(0,\dots,0)\Big) \end{align}$$

This fact can be proved by iterated integration and taylor expansion around $(0,\dots ,0)$.

Then, $$\begin{align} \Big(\bigstar\Big) =&\sum_{j=1}^p(-1)^{j+1}\frac{\partial}{\partial t_j}f(\sigma^t)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{j=1}^{p+1}(-1)^{j+1}\Big( \sum_{k=1}^n\frac{\partial f}{\partial x_k}\frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ +&\sum_{j=1}^{p+1}(-1)^{j+1}f(\sigma^t)\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{k=1}^n\frac{\partial f}{\partial x_k}\sum_{j=1}^{p+1}(-1)^{j-1}\Big( \frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ +&f(\sigma^t)\sum_{j=1}^{p+1}(-1)^{j-1}\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =\big(&\bigstar\bigstar\big) \end{align}$$

To continue the calculation further more, I will use two formula.

1. $$\begin{align}&\sum_{j=1}^{p+1}(-1)^{j-1}\Big( \frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&det \begin{vmatrix} \frac{\sigma^t_k}{\partial t_1} & \cdots & \frac{\sigma^t_k}{\partial t_{j-1}} &\frac{\sigma^t_k}{\partial t_{j}}&\frac{\sigma^t_k}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_k}{\partial t_{p+1}}\\ \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} & \frac{\sigma^t_1}{\partial t_{j}}&\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}} & \frac{\sigma^t_l}{\partial t_{j}} & \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} & \frac{\sigma^t_{p}}{\partial t_{j}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&dx_k\wedge dx_1\wedge dx_2\wedge\dots\wedge dx_p (\frac{\sigma^t}{\partial t_{1}},\frac{\sigma^t}{\partial t_{2}},\dots,\frac{\sigma^t}{\partial t_{p+1}})_{\Big\lvert_{t_i=0}} \end{align}$$

2. $$\begin{align} &\sum_{j=1}^{p+1}(-1)^{j-1}\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{\tau\in S_{p+1}} \big(sgn(\tau)\big) \frac{\sigma^t_{1}}{\partial t_{\tau(1)}\partial t_{\tau(2)}} \frac{\sigma^t_{2}}{\partial t_{\tau(3)}} \frac{\sigma^t_{3}}{\partial t_{\tau(4)}}\dots \frac{\sigma^t_{p}}{\partial t_{\tau(p+1)}} _{\Big\lvert_{t_i=0}}\\ =&0 \\ &\big( \text{since}\quad sgn(\tau)\frac{\sigma^t_{1}}{\partial t_{\tau(1)}\partial t_{\tau(2)}} =-sgn(\tau{'})\frac{\sigma^t_{1}}{\partial t_{\tau{'}(1)}\partial t_{\tau{'}(2)}} \quad \text{where}\quad \tau(1)=\tau^{'}(2),\tau(l)=\tau^{'}(l)\text{for} \quad l\geq3\big) \end{align}$$

Applying 1.and 2.to$\big(\bigstar\bigstar\big)$,I finally get $$ \begin{align} &\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma^t}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\sum_{k=1}^n\frac{\partial f}{\partial x_k} dx_k\wedge dx_1\wedge dx_2\wedge\dots\wedge dx_p (\frac{\sigma^t}{\partial t_{1}},\frac{\sigma^t}{\partial t_{2}},\dots,\frac{\sigma^t}{\partial t_{p+1}})_{\Big\lvert_{t_i=0}} \end{align} $$

Then when $\sigma^t(t_1,t_2,\dots ,t_{p+1})=\sum_i t_iv_i\in U$, and $\sigma^t(0)=(x_1,x_2,\dots,x_n)$,this is what I needed first.

Exterior derivative of differential p-form $\omega$ can be defined by "(p+1)-linear part of the value of $\omega$ integrated over the boundary of infinitesimal (p+1)-parallelotope".

More specifically, $$d\omega(v_1,v_2,...,v_{p+1})\\ =\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}\omega$$

where $[tv_1,tv_2,...,tv_{p+1}]$ is (p+1)-parallelotope spanned by $tv_1,tv_2,...,tv_{p+1}$.

This aspect of exterior derivative is already mentioned by Petya and MathCrawler, but there is no proof why it is equal to the standard definition of exterior derivative . So I'll give you.

It suffices to prove that

$$d\big(f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\big)(v_1,v_2,...,v_{p+1})\\= \sum_{i\in \{ 1,2,3,...,n \} }\frac{f(x_1,x_2,...,x_n)}{\partial x_i} x_i\wedge x_1\wedge x_2\wedge ...\wedge x_p (v_1,v_2,...,v_{p+1})$$

is equal to $$\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial[tv_1,tv_2,...,tv_{p+1}]}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$$

Suppose $U\subset \mathbb{R^n} $ is open, $\sigma^t:[0,t]^{p+1}\to U$ is $C^{\infty}$and, $\mathbf{t} \mapsto \big(\sigma^t_1(\mathbf{t}),\sigma^t_2(\mathbf{t}),\sigma^t_3(\mathbf{t}),...,\sigma^t_n(\mathbf{t})\big)\in U$ $f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p$ is differential p-form on $U$

In order to define $\partial\sigma^t$ with induced orientation, let $d^j_-(t_1, \dots, t_{p+1})=(t_1, \dots, t_{j-1}, 0, t_{j+1}, \dots, t_{p+1}),\\ d^j_+(t_1, \dots, t_{p-1}) = (t_1, \dots, t_{j-1}, t, t_{j+1}, \dots, t_{p+1}).$

Then $\partial \sigma^t = \sum_{j=1}^{p+1} (-1)^j(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)$, and it can be computed as below

$$ \begin{align} &\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma^t}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^{p+1}(-1)^j\int_{(\sigma^t \circ d^j_- - \sigma^t \circ d^j_+)}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^{p+1}(-1)^j\Big(\int_{\sigma^t\circ d^j_-} f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p -\int_{\sigma^t\circ d^j_+} f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\Big)\\ =&\lim_{t\to0}\frac1{t^{p+1}}\sum_{j=1}^p(-1)^j\Big(\int_{[0,t]^p}f(\sigma^t\circ d^j_-)det \begin{vmatrix} \frac{\sigma^t_1\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_-}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_-}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}\circ d^j_-}{\partial t_1} &\cdots & \frac{\sigma^t_l\circ d^j_-}{\partial t_{j-1}}& \frac{\sigma^t_l\circ d^j_-}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l\circ d^j_-}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}\circ d^j_-}{\partial t_1} & \cdots & \frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j-1}} &\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_-}{\partial t_{p+1}} \end{vmatrix}dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p -\int_{[0,t]^p}f(\sigma^t\circ d^j_+)det \begin{vmatrix} \frac{\sigma^t_1\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_1\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_1\circ d^j_+}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1\circ d^j_+}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}\circ d^j_+}{\partial t_1} &\cdots & \frac{\sigma^t_l\circ d^j_+}{\partial t_{j-1}}& \frac{\sigma^t_l\circ d^j_+}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_k\circ d^j_+}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}\circ d^j_+}{\partial t_1} & \cdots & \frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j-1}} &\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}\circ d^j_+}{\partial t_{p+1}} \end{vmatrix} dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\Big)\\ \Big(=&\bigstar\Big) \end{align} $$

To Continue the computation, I will use a following fact.

$$\begin{align} &\lim_{t\to 0}\frac1{t^{p+1}}\int_{[0,t]^p}g(d^t_+)-g(d^t_-) dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\\ =&\lim_{t\to 0}\frac1{t^{p+1}}\int_{[0,t]^p}g(t_1,\dots ,t_{j-1},t,t_{j+1},\dots,t_{p+1})-g(t_1,\dots ,t_{j-1},0,t_{j+1},\dots,t_{p+1}) dt_1\dots dt_{j-1}dt_{j+1}\dots dt_p\\ =&\frac{\partial}{\partial t_j}g(t_1,\dots ,t_j,\dots,t_{p+1})\rvert_{t_{i=0}}\quad \Big(=\frac{\partial g}{\partial t_j}(0,\dots,0)\Big) \end{align}$$

This fact can be proved by iterated integration and taylor expansion around $(0,\dots ,0)$.

Then, $$\begin{align} \Big(\bigstar\Big) =&\sum_{j=1}^p(-1)^{j+1}\frac{\partial}{\partial t_j}f(\sigma^t)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{j=1}^{p+1}(-1)^{j+1}\Big( \sum_{k=1}^n\frac{\partial f}{\partial x_k}\frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ +&\sum_{j=1}^{p+1}(-1)^{j+1}f(\sigma^t)\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{k=1}^n\frac{\partial f}{\partial x_k}\sum_{j=1}^{p+1}(-1)^{j-1}\Big( \frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ +&f(\sigma^t)\sum_{j=1}^{p+1}(-1)^{j-1}\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =\big(&\bigstar\bigstar\big) \end{align}$$

To continue the calculation further more, I will use two formula.

1. $$\begin{align}&\sum_{j=1}^{p+1}(-1)^{j-1}\Big( \frac{\sigma^t_k}{\partial t_j}\Big)det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&det \begin{vmatrix} \frac{\sigma^t_k}{\partial t_1} & \cdots & \frac{\sigma^t_k}{\partial t_{j-1}} &\frac{\sigma^t_k}{\partial t_{j}}&\frac{\sigma^t_k}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_k}{\partial t_{p+1}}\\ \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} & \frac{\sigma^t_1}{\partial t_{j}}&\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}} & \frac{\sigma^t_l}{\partial t_{j}} & \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} & \frac{\sigma^t_{p}}{\partial t_{j}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&dx_k\wedge dx_1\wedge dx_2\wedge\dots\wedge dx_p (\frac{\sigma^t}{\partial t_{1}},\frac{\sigma^t}{\partial t_{2}},\dots,\frac{\sigma^t}{\partial t_{p+1}})_{\Big\lvert_{t_i=0}} \end{align}$$

2. $$\begin{align} &\sum_{j=1}^{p+1}(-1)^{j-1}\frac{\partial}{\partial t_j}det \begin{vmatrix} \frac{\sigma^t_1}{\partial t_1} & \cdots & \frac{\sigma^t_1}{\partial t_{j-1}} &\frac{\sigma^t_1}{\partial t_{j+1}}& \cdots &\frac{\sigma^t_1}{\partial t_{p+1}}\\ \vdots & \ddots & \vdots & \vdots &\ddots&\vdots \\ \frac{\sigma^t_{l}}{\partial t_1} &\cdots & \frac{\sigma^t_l}{\partial t_{j-1}}& \frac{\sigma^t_l}{\partial t_{j+1}} & \cdots &\frac{\sigma^t_l}{\partial t_{p+1}}\\ \vdots & \ddots &\vdots &\vdots & \ddots & \vdots \\ \frac{\sigma^t_{p}}{\partial t_1} & \cdots & \frac{\sigma^t_{p}}{\partial t_{j-1}} &\frac{\sigma^t_{p}}{\partial t_{j+1}} & \cdots&\frac{\sigma^t_{p}}{\partial t_{p+1}} \end{vmatrix}_{\Big\lvert_{t_i=0}}\\ =&\sum_{\tau\in S_{p+1}} \big(sgn(\tau)\big) \frac{\sigma^t_{1}}{\partial t_{\tau(1)}\partial t_{\tau(2)}} \frac{\sigma^t_{2}}{\partial t_{\tau(3)}} \frac{\sigma^t_{3}}{\partial t_{\tau(4)}}\dots \frac{\sigma^t_{p}}{\partial t_{\tau(p+1)}} _{\Big\lvert_{t_i=0}}\\ =&0 \\ &\big( \text{since}\quad sgn(\tau)\frac{\sigma^t_{1}}{\partial t_{\tau(1)}\partial t_{\tau(2)}} =-sgn(\tau{'})\frac{\sigma^t_{1}}{\partial t_{\tau{'}(1)}\partial t_{\tau{'}(2)}} \quad \text{where}\quad \tau(1)=\tau^{'}(2),\tau(l)=\tau^{'}(l)\text{for} \quad l\geq3\big) \end{align}$$

Applying 1.and 2.to$\big(\bigstar\bigstar\big)$,I finally get $$ \begin{align} &\lim_{t\to0}\frac1{t^{p+1}}\int_{\partial\sigma^t}f(x_1,x_2,...,x_n)x_1\wedge x_2\wedge ...\wedge x_p\\ =&\sum_{k=1}^n\frac{\partial f}{\partial x_k} dx_k\wedge dx_1\wedge dx_2\wedge\dots\wedge dx_p (\frac{\sigma^t}{\partial t_{1}},\frac{\sigma^t}{\partial t_{2}},\dots,\frac{\sigma^t}{\partial t_{p+1}})_{\Big\lvert_{t_i=0}} \end{align} $$

Then when $\sigma^t(t_1,t_2,\dots ,t_{p+1})=\sum_i t_iv_i\in U$, and $\sigma^t(0)=(x_1,x_2,\dots,x_n)$,this is what I needed first.