Analytically, I want to prove the following :
Let $\mathbb{F}_Q$ be an extension field of $\mathbb{F}_q$. Consider the polynomial $g(x)=x^q-x$ over $\mathbb{F}_Q[x]$. Show that the roots of $g(x)$ are the elements of $\mathbb{F}_q$.
CLAIM 1: $g(x)=x^q-x$, $q\geq2$ has at most $q$ roots in every extension field of $\mathbb{F}_q$.
Proof: (by degree counting) i.e. let $\beta_1,\beta_2,\dots$ be the roots of $g(x)$ in the extension field $\mathbb{F}_Q$ of $\mathbb{F}_q$ and let $m_i$ be the multiplicity of root $\beta_i$. Then $$\mathrm{degree}(\prod\limits_i(x-\beta_i)^{m_i})=\sum\limits_im_i \text{\quad and\quad} \prod\limits_i(x-\beta_i)^{m_i}\vert g(x)\Rightarrow\sum\limits_im_i\leq q$$
CLAIM 2: $\prod\limits_{\beta\in \mathbb{F}_q}(x-\beta)=x^{|\mathbb{F}_q|}-x$ (1)
Proof: By Lagrange's Theorem every element of $\mathbb{F}_q$ is a root of $x^{|\mathbb{F}_q|}-x\Rightarrow x-\beta\vert x^{|\mathbb{F}_q|}-x, \forall \beta\in \mathbb{F}_q$ (CLAIM 3).
Proof of CLAIM 3: if $g(x)$ is a polynomial over $F_q$ and $\beta$ an element of an extension field $\mathbb{F}_Q$ of $\mathbb{F}_q$ then $g(\beta)=0$ iff $x-\beta\vert g(x)$ over $\mathbb{F}_Q[x]$ (CLAIM 4).
Proof of CLAIM 4: $g(x)=q(x)(x-\beta)+r(x)$, $\mathrm{degree}(r(x))<1$. Then $g(\beta)=0$ iff $r(x)=0$.
Now, CLAIM 2 follows since in (1) $\mathrm{degree(LHS)}=\mathrm{degree(RHS)}$ and both sides are monic. Finally, since $\mathbb{F}_q\subseteq \mathbb{F}_Q$ we can see that the roots of $g(x)$ over $\mathbb{F}_Q[x]$ are exactly the elements of $\mathbb{F}_q$.
Do you think this derivation is correct?