The Frobenius map is defined as follows:
${\Phi}: GF(p^n) \rightarrow GF(p^n)$
where ${\Phi}(x)= x^p$
I already showed that the map is a homomorphism and is 1-1. I am left to show that it is onto.
Here is what I have so far:
Let $y \in GF(p^n) $ such that there does not exist any $x \in GF(p^n) $ that satisfies $y=x^p$
I'm not sure how to go on from here. What I know is $GF(p^n)$ contains all roots of $x^{p^n} -x$ but I don't know how to use it for the proof.