 Solution

For $\alpha \ne 0$, Lagrange's theorem in the multiplicative group of order $q^m-1$ gives ${\alpha }^{q^m-1}=1$; at zero the desired identity is immediate. Thus all elements satisfy ${\alpha }^{q^m}-\alpha =0$.

For an irreducible degree-$m$ polynomial $f\in F[X]$, the quotient $L=F[X]/(f)$ is a field of dimension $m$, hence has $q^m$ elements by part 1. Its element $\alpha =[X]$ therefore satisfies ${\alpha }^{q^m}-\alpha =0$. The evaluation homomorphism $F[X]\to L$ has kernel $(f)$, so