This can be proved by induction on $k=0,1,2,\dots$. Indeed, let $g_0(f;x)=f(x)$, $g_n(f;x)=g_{n-1}(f;x+h_n)-g_{n-1}(f;x)$ for $n=1,2,\dots$.

We want to show that for all natural $k$ and all $x$ such that $a\le x\le x+h_1+\cdots+h_k\le b$ there is $c_k=c_k(f;x)$ such that $a<c_k\le c_k+h_1+\cdots+h_{k-1}<b$ and $$g_k(f;x) = f^{(k)}(c_k)h_1\cdots h_k. \tag{1} $$

For $k=0$, this is trivial. Suppose this is true for $k=n$, where $n\in\{0,1,\dots\}$. Then, by the mean value theorem, for all $x$ such that $a\le x\le x+h_1+\cdots+h_{n+1}\le b$ we have \begin{align} g_{n+1}(f;x)&=g_n(f;x+h_{n+1})-g_n(f;x)\\ &=g_n'(f;b_n)h_{n+1}\\ &=g_n(f';b_n)h_{n+1} \\ &={f'}^{(n)}(c_{n+1})h_1\cdots h_nh_{n+1} \\ &=f^{(n+1)}(c_{n+1})h_1\cdots h_{n+1} \end{align} for some $b_n=b_n(f;x)$ such that $x<b_n<x+h_{n+1}$ and $c_{n+1}:=c_n(f';b_n)$, so that $a<c_{n+1}\le c_{n+1}+h_1+\cdots+h_n<b$, as desired.

The third equality in the above multi-line display is the crucial (even if very simple) observation, and the fourth equality in that display follows by induction, because the conditions $a\le x\le x+h_1+\cdots+h_{n+1}\le b$ and $x<b_n<x+h_{n+1}$ imply $a\le x<b_n\le b_n+h_1+\cdots+h_n<x+h_1+\cdots+h_{n+1}\le b$, so that $a\le b_n\le b_n+h_1+\cdots+h_n\le b$.

This can be proved by induction on $k=0,1,2,\dots$. Indeed, let $g_0(f;x)=f(x)$, $g_n(f;x)=g_{n-1}(f;x+h_n)-g_{n-1}(f;x)$ for $n=1,2,\dots$.

We want to show that for all $k=0,1\dots$ and all $x$ such that $a\le x\le x+h_1+\cdots+h_k\le b$ there is $c_k=c_k(f;x)$ such that $a<c_k\le c_k+h_1+\cdots+h_{k-1}<b$ and $$g_k(f;x) = f^{(k)}(c_k)h_1\cdots h_k. \tag{1} $$

For $k=0$, this is trivial. Suppose this is true for $k=n$, where $n\in\{0,1,\dots\}$. Then, by the mean value theorem, for all $x$ such that $a\le x\le x+h_1+\cdots+h_{n+1}\le b$ we have \begin{align} g_{n+1}(f;x)&=g_n(f;x+h_{n+1})-g_n(f;x)\\ &=g_n'(f;b_n)h_{n+1}\\ &=g_n(f';b_n)h_{n+1} \\ &={f'}^{(n)}(c_{n+1})h_1\cdots h_nh_{n+1} \\ &=f^{(n+1)}(c_{n+1})h_1\cdots h_{n+1} \end{align} for some $b_n=b_n(f;x)$ such that $x<b_n<x+h_{n+1}$ and $c_{n+1}:=c_n(f';b_n)$, so that $a<c_{n+1}\le c_{n+1}+h_1+\cdots+h_n<b$, as desired.

The third equality in the above multi-line display is the crucial (even if very simple) observation, and the fourth equality in that display follows by induction, because the conditions $a\le x\le x+h_1+\cdots+h_{n+1}\le b$ and $x<b_n<x+h_{n+1}$ imply $a\le x<b_n\le b_n+h_1+\cdots+h_n<x+h_1+\cdots+h_{n+1}\le b$, so that $a\le b_n\le b_n+h_1+\cdots+h_n\le b$.

This can be proved by induction on $k=0,1,2,\dots$. Indeed, let $g_0(f;x)=f(x)$, $g_n(f;x)=g_{n-1}(f;x+h_n)-g_{n-1}(f;x)$ for $n=1,2,\dots$.

We want to show that for all natural $k$ and all $x$ such that $a\le x\le x+h_1+\cdots+h_k\le b$ there is $c_k=c_k(f;x)$ such that $a<c_k\le c_k+h_1+\cdots+h_{k-1}<b$ and $$g_k(f;x) = f^{(k)}(c_k)h_1\cdots h_k. \tag{1} $$

For $k=0$, this is trivial. Suppose this is true for $k=n$, where $n\in\{0,1,\dots\}$. Then, by the mean value theorem, for all $x$ such that $a\le x\le x+h_1+\cdots+h_{n+1}\le b$ we have \begin{align} g_{n+1}(f;x)&=g_n(f;x+h_{n+1})-g_n(f;x)\\ &=g_n'(f;b_n)h_{n+1}\\ &=g_n(f';b_n)h_{n+1} \\ &={f'}^{(n)}(c_{n+1})h_1\cdots h_nh_{n+1} \\ &=f^{(n+1)}(c_{n+1})h_1\cdots h_{n+1} \end{align} for some $b_n:=c_n(f';x)$ such that $x<b_n<x+h_{n+1}$ and some $c_{n+1}$ such that $a<c_{n+1}\le c_{n+1}+h_1+\cdots+h_n<b$, as desired.

The third equality in the above multi-line display is the crucial (even if very simple) observation, and the fourth equality in that display follows by induction, because the conditions $a\le x\le x+h_1+\cdots+h_{n+1}\le b$ and $x<b_n<x+h_{n+1}$ imply $a\le x<b_n\le b_n+h_1+\cdots+h_n<x+h_1+\cdots+h_{n+1}\le b$, so that $a\le b_n\le b_n+h_1+\cdots+h_n\le b$.

This can be proved by induction on $k=0,1,2,\dots$. Indeed, let $g_0(f;x)=f(x)$, $g_n(f;x)=g_{n-1}(f;x+h_n)-g_{n-1}(f;x)$ for $n=1,2,\dots$.

We want to show that for all natural $k$ and all $x$ such that $a\le x\le x+h_1+\cdots+h_k\le b$ there is $c_k=c_k(f;x)$ such that $a<c_k\le c_k+h_1+\cdots+h_{k-1}<b$ and $$g_k(f;x) = f^{(k)}(c_k)h_1\cdots h_k. \tag{1} $$

For $k=0$, this is trivial. Suppose this is true for $k=n$, where $n\in\{0,1,\dots\}$. Then, by the mean value theorem, for all $x$ such that $a\le x\le x+h_1+\cdots+h_{n+1}\le b$ we have \begin{align} g_{n+1}(f;x)&=g_n(f;x+h_{n+1})-g_n(f;x)\\ &=g_n'(f;b_n)h_{n+1}\\ &=g_n(f';b_n)h_{n+1} \\ &={f'}^{(n)}(c_{n+1})h_1\cdots h_nh_{n+1} \\ &=f^{(n+1)}(c_{n+1})h_1\cdots h_{n+1} \end{align} for some $b_n=b_n(f;x)$ such that $x<b_n<x+h_{n+1}$ and $c_{n+1}:=c_n(f';b_n)$, so that $a<c_{n+1}\le c_{n+1}+h_1+\cdots+h_n<b$, as desired.

The third equality in the above multi-line display is the crucial (even if very simple) observation, and the fourth equality in that display follows by induction, because the conditions $a\le x\le x+h_1+\cdots+h_{n+1}\le b$ and $x<b_n<x+h_{n+1}$ imply $a\le x<b_n\le b_n+h_1+\cdots+h_n<x+h_1+\cdots+h_{n+1}\le b$, so that $a\le b_n\le b_n+h_1+\cdots+h_n\le b$.