Problem
Let $F$ be a finite field with $q$ elements. Let $f\in F[x]$ be an irreducible polynomial. Prove that if $f \mid x^{q^n}-x$ then $\deg{f}\mid n$ (the converse is also true and I have a proof).
Attempt
Since $f\mid x^{q^n}-x$, we have the following tower of extensions: $\mathbb{F}_{q^n}/L/F(\alpha)/F$, where $\mathbb{F}_{q^n}$ is the splitting field of $x^{q^n}-x$, $L$ is the splitting field of $f$, and $F(\alpha)$ is the field extension of $F$ obtained by adjoining a root of $f$.
So $\deg f=[F(\alpha):F]\mid [\mathbb{F}_{q^n}:F]=n$.
A proof of the converse
Suppose that $d=\deg f\mid n$. Then we have that $f\mid x(x^{q^{d}-1}-1)\mid x(x^{q^{n}-1}-1)$:
(i) $f\mid x(x^{q^{d}-1}-1)$: for any root $\alpha$ of $f$, $\alpha^{q^d}=\alpha$ because $|F(\alpha)^*|=q^d-1$. Then $f\mid(x^{q^{d}}-x) $ since $f$ is the minimal polynomial of $\alpha$ over $F$.
(ii) $(x^{q^{d}-1}-1)\mid (x^{q^{n}-1}-1)$: Since $d\mid n$, we have $q^n-1=(q^d-1)[(q^{d})^{n/d-1}+\dots+q^d+1]$. So $(q^d-1)\mid (q^n-1)$. Applying this trick again we get that $(x^{q^{d}-1}-1)\mid (x^{q^{n}-1}-1)$.
Answered by the original poster.