 Solution

For $\alpha \ne 0$, 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 $1,\overline{X},\dots ,{\overline{X}}^{m-1}$. 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