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Consider the map $\varphi:\mathbb{F}_{p^{n}} \rightarrow \mathbb{F}_{p^{n}}$ defined by $x \mapsto x^{p}-x$. Since $(a+b)^{p}= a^{p}+b^{p}$ for all $a,b \in \mathbb{F}_{p^{n}}$ we have that $\varphi$ is a homomorphism. An element of $\mathbb{F}_{p^{n}}$ can be written in the form $a^{p}-a$ iff it is in Im$(\varphi)$, so it suffices to find $|$Im$(\varphi)|$, which is equal to $\mathbb{F}_{p^{n}}/K$ where $K=$ker$\varphi=\{a \in \mathbb{F}_{p^{n}} : a^{p}-a = 0\}$. So in particular, if $a \in$ ker$\varphi$ then $a$ is a root of the polynomial $x^{p}-x$ which is a separable polynomial so has precisely $p$ roots. Since $a^{p}=a$ for all $a \in \mathbb{F}_{p} \subseteq \mathbb{F}_{p^{n}}$, which is all $p$ of the roots, we have $|$ker$\varphi| = p$. Hence $|$Im$(\varphi)|=p^{n-1}$ so this is the number of such elements.

I am pretty sure this works, but is there an easier or shorter way?

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Your method is essentially how I do it when teaching this bit (and how it is done in the literature). Well done.

A related result that is often given in this context is the following description of the image.

$$\operatorname{im}(\phi)=\{a\in\Bbb{F}_{p^n}\mid tr(a)=0\}=\operatorname{ker}(tr),$$

where

$$tr(x)=x+x^p+x^{p^2}+\cdots+x^{p^{n-1}}$$

is the trace map $tr:\Bbb{F}_{p^n}\to \Bbb{F}_p$

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