$x^{p^n}-x$ is the product of monic irreducibles in $\FF_p[x]$ of degree dividing $n$

Let \[ P(x)=x^{p^n}-x\in\FF_p[x]. \] Its derivative is \[ P'(x)=p^n x^{p^n-1}-1=-1, \] so \(P\) is squarefree.

Let \(f\in\FF_p[x]\) be monic irreducible of degree \(d\), and let \(\alpha\) be a root. Then \[ \FF_p(\alpha)\cong\FF_{p^d}. \] Now \[ f\mid P \iff \alpha^{p^n}=\alpha \iff \alpha\in\FF_{p^n} \iff \FF_{p^d}\subseteq\FF_{p^n} \iff d\mid n. \] Thus the irreducible factors of \(P\) are exactly the monic irreducibles whose degrees divide \(n\). Since \(P\) is monic and squarefree, \[ \boxed{x^{p^n}-x=\prod_{d\mid n}\ \prod_{\substack{f\text{ monic irreducible}\\ \deg f=d}} f(x).} \]