= Solution
For $\alpha\neq0$, <Lagrange's theorem for finite groups> in $K^\times$, whose order is $q^m-1$, gives $\alpha^{q^m-1}=1$. The zero element satisfies the equation too, so every $\alpha\in K$ has $\alpha^{q^m}=\alpha$.
Now let $f\in F[X]$ be irreducible of degree $m$. The quotient $F[X]/(f)$ is a <field> with $q^m$ elements, by its <basis> $1,\overline X,\ldots,\overline X^{m-1}$. Its element $\overline X$ therefore satisfies $\overline X^{q^m}-\overline X=0$. This means precisely that $X^{q^m}-X$ lies in the ideal $(f)$, giving
$$
\boxed{f(X)\mid X^{q^m}-X.}
$$
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