Let \(p\) be prime and \(n\geq 1\), and set \(\rho_n(x) \coloneqq x^{p^n} - x\). Then \[ \rho_n(x) = \prod_{\substack{f\in\FF_p[x]\text{ monic irreducible}\\ \deg f\mid n}} f(x). \] Equivalently, for a monic irreducible \(f\in\FF_p[x]\), \[ f(x)\mid \rho_n(x) \iff \deg f\mid n. \]
The roots of \(\rho_n(x)=x^{p^n}-x\) in an algebraic closure of \(\FF_p\) are exactly the elements of \(\FF_{p^n}\). If \(f\) is monic irreducible of degree \(d\) and \(\alpha\) is one of its roots, then \(\FF_p(\alpha)\cong\FF_{p^d}\). Hence \[ f\mid \rho_n \iff \alpha\in\FF_{p^n} \iff \FF_{p^d}\subseteq\FF_{p^n} \iff d\mid n. \] Finally, \(\rho_n'(x)=-1\), so \(\rho_n\) is squarefree; therefore each of these irreducible factors occurs exactly once, giving the displayed product.