Proving that a Finite Field Over Its Prime Field Is Galois

Question

In the above, I don't understand why the author needs to use another theorem to show that $f$ is separable. Theorem 3.4.5 says that in a finite field, say $F$, the Frobenius automorphism gives us $F = F^p$, entailing that every irreducible polynomial in $F[x]$ is separable. My thinking is, if $E$ exhausts all possible $p^n$ distinct roots of $f$, then $f$ just cannot have repeated roots. This is pretty much the last statement in the paragraph. So, was the author being redundant?

Answer 1

You're correct. This is not really needed and you know right away that f is separable. In general, the idea you have outlined is a common and excellent way to show that a polynomial is separable.