Here is a proof for $\dim(V^*)>\dim(V)$ for every infinite dimensional vector space $V$ over a field $k$. More precisely, we prove that $$\mathrm{dim}(V^*)=|V^*|=|k|^{\dim(V)}\geq 2^{\dim(V)}>\dim(V).$$

We first show that $\dim(V^*)\geq |k|$. Fix a basis $B$ for $V$ and choose a countable infinite subset $v_0,v_1,v_2,\ldots$ of vectors of $B$. Given $\alpha\in k$, we define $f_\alpha\in V^*$ to be the unique functional defined on the basis elements by setting $f_\alpha(v_n):=\alpha^n$ for all $n=0,1,2,\ldots$ and $f_\alpha(v)=0$ for all other elements $v\in B$. Notice that $(f_\alpha)_{\alpha\in k}$ is a family of pairwise distinct functionals of $V$.

We claim that the family $(f_\alpha)_{\alpha\in k}$ is linearly independent. Indeed, if $\alpha_1,\ldots,\alpha_n\in k$ are distinct elements and $$x_1f_{\alpha_1}+\ldots+x_nf_{\alpha_n}=0$$ for certain scalars $x_i\in k$, then evaluating at $v_0, v_1,\ldots, v_{n-1}$ we get a system of $n$ linear equations of the form $$x_1\alpha_1^i+\ldots+x_n\alpha_n^i=0,\quad i=0,1,\ldots,n-1$$ in the variables $x_1,\ldots,x_n$. The $n\times n$ matrix of this system is $$\left(\begin{array}{ccc} 1& 1 &\ldots & 1\\ \alpha_1& \alpha_2 &\ldots & \alpha_n\\ \alpha_1^2& \alpha_2^2 &\ldots & \alpha_n^2\\ \vdots & \ldots & \ldots &\vdots \\ \alpha_1^{n-1} &\alpha_2^{n-1} &\ldots &\alpha_n^{n-1} \end{array}\right)$$ This is a (transpose of a) Vandermonde matrix, which is therefore invertible, so that $x_1=x_2=\ldots=x_n=0$, as desired.

It follows that $$|V^*|=\max\{|k|,\dim(V^*)\}=\dim(V^*).$$ Since there is an isomorphism of vector spaces $V^*\cong V^B$ (where the right hand side denotes the space of all functions $B\to V$), it also follows $$\dim(V^*)=|V^B|=|V|^{\dim(V)}.$$

Here is a proof for $\dim(V^*)>\dim(V)$ for every infinite dimensional vector space $V$ over a field $k$. More precisely, we prove that $$\mathrm{dim}(V^*)=|V^*|=|k|^{\dim(V)}\geq 2^{\dim(V)}>\dim(V).$$

We first show that $\dim(V^*)\geq |k|$. Fix a basis $B$ for $V$ and choose a countable infinite subset $v_0,v_1,v_2,\ldots$ of vectors of $B$. Given $\alpha\in k$, we define $f_\alpha\in V^*$ to be the unique functional defined on the basis elements by setting $f_\alpha(v_n):=\alpha^n$ for all $n=0,1,2,\ldots$ and $f_\alpha(v)=0$ for all other elements $v\in B$. Notice that $(f_\alpha)_{\alpha\in k}$ is a family of pairwise distinct functionals of $V$.

We claim that the family $(f_\alpha)_{\alpha\in k}$ is linearly independent. Indeed, if $\alpha_1,\ldots,\alpha_n\in k$ are distinct elements and $$x_1f_{\alpha_1}+\ldots+x_nf_{\alpha_n}=0$$ for certain scalars $x_i\in k$, then evaluating at $v_0, v_1,\ldots, v_{n-1}$ we get a system of $n$ linear equations of the form $$x_1\alpha_1^i+\ldots+x_n\alpha_n^i=0,\quad i=0,1,\ldots,n-1$$ in the variables $x_1,\ldots,x_n$. The $n\times n$ matrix of this system is $$\left(\begin{array}{ccc} 1& 1 &\ldots & 1\\ \alpha_1& \alpha_2 &\ldots & \alpha_n\\ \alpha_1^2& \alpha_2^2 &\ldots & \alpha_n^2\\ \vdots & \ldots & \ldots &\vdots \\ \alpha_1^{n-1} &\alpha_2^{n-1} &\ldots &\alpha_n^{n-1} \end{array}\right)$$ This is a (transpose of a) Vandermonde matrix, which is therefore invertible, so that $x_1=x_2=\ldots=x_n=0$, as desired.

It follows that $$|V^*|=\max\{|k|,\dim(V^*)\}=\dim(V^*).$$ Since there is an isomorphism of vector spaces $V^*\cong k^B$ (where the right hand side denotes the space of all functions $B\to k$), it also follows $$\dim(V^*)=|k^B|=|k|^{\dim(V)}.$$