I’m struggling to understand one part of this proof. I understand that the size of $GF(p^n)$ over $\mathbb{Z}_p$ is $n$ and that the Frobenius automorphism $\Phi$ is an element of the Galois group. I fail to understand why the order of $\Phi$ is $n$.
The order of $\Phi$ divides $n$ by Cauchy’s theorem. The order of $\Phi$ is the smallest $m \in \mathbb{N}$ such that $\Phi ^m = id $ i.e. the smallest $m$ such that $\Phi ^m (x) = id(x) \; \; \forall x \in GF(p^n)$ or equivalently $(x^p)^m = x^{pm}=x$.
We know that $x^{p^n}=x \;\; \forall x \in GF(p^n)$ but I don’t know how to apply this to show m=n.
Let $\text{Fr}:\mathbf{F}_{p^n}\to \mathbf{F}_{p^n},x\mapsto x^p$ be the Frobenius automorphism.
For every $d<n$, $\text{Fr}^d$ is not the identity on $\mathbf{F}_{p^n}$, since $X^{p^d}-X$ has at most $p^d$ roots.
Edit. I will elaborate a bit. If $\text{Fr}^d$ were the identity for $d<n$, then all $x\in \mathbf{F}_{p^n}$ would satisfy $x^{p^d}=x$. Contradiction, since $\mathbf{F}_{p^n}$ contains $p^n$ elements and $X^{p^d}-X$ has at most $p^d$ roots. (This is a result from basic ring theory: a polynomial of degree $m$ over an integral domain as at most $m$ roots.)