Over ${\mathbb{F}}_q$, $x^{q^n}-x$ is the product of all monic irreducibles whose degrees divide $n$

By [L1], write $q=p^r$. Use [L2] to choose a field $E$ of order $p^{rn}=q^n$. By [L3], $E$ is the full root set and a splitting field of $t^{q^n}-t$.

Let $P$ be monic irreducible of degree $d\ge 1$. By [L8] it has a root $\alpha$ in some extension of ${\mathbb{F}}_q$; since $P$ is monic irreducible and annihilates $\alpha$, the uniqueness in [L9] makes $P$ the minimal polynomial of $\alpha$. By [L5], ${\mathbb{F}}_q(\alpha )$ has degree $d$ over ${\mathbb{F}}_q$ and hence has $q^d=p^{rd}$ elements.

The derivative of $t^{q^n}-t$ is $-1$, which vanishes nowhere. Every irreducible factor of $t^{q^n}-t$ has a root in the splitting field $E$ of step 1.1, and a repeated factor would make that root repeated; so [L7] shows that no irreducible factor repeats.

If $d\mid n$, write $n=ed$. Applied to the field ${\mathbb{F}}_q(\alpha )$, [L3] gives ${\alpha }^{q^d}=\alpha$; iterating this identity $e$ times gives ${\alpha }^{q^n}=\alpha$. So $t^{q^n}-t$ vanishes at $\alpha$, and since $P$ is its minimal polynomial by step 1.2, [L9] gives $P\mid t^{q^n}-t$.

Conversely, if $P$ divides $t^{q^n}-t$, choose its root $\alpha$ in the splitting field $E$ from step 1.1. Then ${\mathbb{F}}_q(\alpha )$ is a subfield of $E$ with order $p^{rd}$, so [L4] gives $rd\mid rn$, and cancellation yields $d\mid n$.

Factor the polynomial by [L6]. Steps 2.2 and 2.3 identify exactly the monic irreducible factors, and step 2.1 gives multiplicity one. Since both sides are monic, their unit factors agree, proving the formula.