Proving that $f(x)$ divides $x^{p^n} - x$ iff $\deg f(x)$ divides $n$

Hints:

(i) Show that the splitting field of the polynomial $\,p(x):=x^{p^n}-x\in\Bbb F_p[x]\,$ over the prime field $\,\Bbb F_p\,$ is the field $\,\Bbb F_{p^n}\,$

(ii) One way to go: show that the set of roots of the above polynomial $\,p(x)\,$ in some algebraic closure of $\,\Bbb F_p\,$ is a field...

(iii) Prove that $\,\Bbb F_{p^d}\,$ is a subfield of $\,\Bbb F_{p^n}\,$ iff $\,d\mid n\,$

Of course, take into account that $\,f(x)\mid p(x)\Longrightarrow\,$ all the roots of $\,f(x)\,$ are also roots of $\,p(x)\,$

This question is very old, but there is a more direct solution worth noting.

Proof.

Suppose $f(x)$ divides $h(x):=x^{p^n} -x$. Then since $h(x)$ splits over $\mathbb{F}_{p^n}$, so does $f(x)$. Let $\alpha \in \mathbb{F}_{p^n}$ be a root of $f(x)$. Then $\mathbb{F}_{p}(\alpha)\subset \mathbb{F}_{p^n}$, and $[\mathbb{F}_p(\alpha): \mathbb{F}_p] = d$. Finally, we have that $n = [\mathbb{F}_{p^n}: \mathbb{F}_p]= [\mathbb{F}_{p^n}: \mathbb{F}_p(\alpha)][\mathbb{F}_p(\alpha): \mathbb{F}_p]$, completing the proof that $d | n$.