Here is the problem I am attempting:
Suppose that $K$ is a finite field with $|K| = q$. Show that if $f(x) \in K[x]$ is irreducible, then $f$ divides $g(x) = x^{q^n}-x$ in $K[x]$ if and only if $\deg(f)$ divides $n$.
For the => direction, I know that $g(x) = f(x)q(x)$ for some polynomial $q(x)$, and $q^n = deg(g) = deg(f) * deg(q)$, but that doesn't seem helpful as it simply implies that either $deg(f)$, $deg(q)$, or both are a power of $q$.
For the other direction (<=), all I can come up with is that the quotient field $K[x]/<f(x)>$ should have degree $deg(f)$ over $K$, so it also divides $n$, and my instinct is to try and find a way to show that $g(x) = 1$ in the quotient field, but I don't know how to get there from where I am.
Hints:
Let $\;L\;$ be an extension field of $\;K\;$ s.t. $\;\dim_KL=n\;$ . Let us denote by $\;\overline K\;$ some (the) algebraic closure of $\;K\;$ :
(i) Prove that $\;L:=\left\{\;a\in\overline K\;;\;a^{q^n}-a=0\;\right\}\;$ , i.e. $\;L\;$ is the splitting field of the polynomial $\;g(x):=x^{q^n}-x\in K[x]\;$
(ii) Deduce then that $\;f(x)\mid g(x)\iff\;$ all the roots of $\;f(x)\;$ are in $\;L\;$ .
(iii) Finally, take now the splitting field of $\;f(x)\;$ over $\;K\;$ and complete the solution.