Every Galois field $F$ of characteristic $p$ is perfect

Question

I'm trying to do Exercise 2.6.13 from textbook Groups, Matrices, and Vector Spaces - A Group Theoretic Approach to Linear Algebra by James B. Carrell. Could you please confirm if my attempt is fine or contains logical mistakes?

Every Galois field $F$ of characteristic $p$ is perfect, i.e. every element is a $p$th power.

My attempt:

We need a lemma from previous exercise

Suppose $F$ is a field of characteristic $p$. If $a, b \in F$ and $a^{p}=b^{p}$, then $a=b$.

Consider a map $f: F \to F, x \mapsto x^p$. By our lemma, $f$ is injective. Moreover, $F$ is finite. Hence any $y \in F$ can be written as $y = x^p$ for some $p \in F$. This completes the proof.

Answer 1

Here is @Mummy the turkey's comment that answers my question. I post it here to remove this question from unanswered list. All credits are given to @Mummy the turkey.

Your proof looks good! Maybe it would be fun to prove why "every element is $p^{th}$ power" implies "irreducible polynomials have distinct roots" which is the usual definition of perfect.