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

Let \(f\in\FF_p[x]\) be monic irreducible of degree \(d\), and let \(\alpha\) be one of its roots. Then \[ \FF_p(\alpha)\cong\FF_{p^d}. \] Hence \[ f\mid x^{p^n}-x \] if and only if \(\alpha\in\FF_{p^n}\), equivalently \[ \FF_{p^d}\subseteq\FF_{p^n}. \] The subfield criterion for finite fields gives \[ \FF_{p^d}\subseteq\FF_{p^n}\iff d\mid n. \] Thus the irreducible factors are exactly the monic irreducibles whose degrees divide \(n\).

Finally, \[ \frac{d}{dx}(x^{p^n}-x)=-1, \] so \(x^{p^n}-x\) is squarefree. Therefore every such irreducible occurs with multiplicity one, and \[ x^{p^n}-x =\prod_{\substack{f\text{ monic irreducible}\\ \deg f\mid n}} f(x). \]