 Solution

Write $Q=q^m$ and take a splitting field $L$ of $X^Q-X$ over $F$. Its formal derivative is $-1$, since the characteristic divides $Q$, so it has $Q$ distinct roots in $L$. Let $K$ be their set. The Frobenius endomorphism identity givesThus $K$ contains $0,1$ and is closed under addition, negatives and multiplication. If $u\ne 0$ belongs to $K$, then $(u^{-1}{)}^Q=(u^Q{)}^{-1}=u^{-1}$, so it is closed under inverses too. Hence $K$ is a field with exactly $Q$ elements. Every $c\in F$ satisfies $c^q=c$ by Lagrange's theorem for finite groups applied to $F^{\times }$, together with the zero case; iterating gives $c^Q=c$. Therefore