Is $\mathbb{F}_{p^r}/\mathbb{F}_{p}$ a Galois extension? Let $r>0$. Is $\mathbb{F}_{p^r}/\mathbb{F}_{p}$ a Galois extension? If so, why?
I know that it is a finite extension, with $[\mathbb{F}_{p^r}:\mathbb{F}_{p}]=r$. To show that it is a Galois extension, it suffices to show that $|Aut(\mathbb{F}_{p^r}/\mathbb{F}_{p})|=r$. 
The notaion $Aut(K/F)$ indicates the group of field automorphisms $K\to K$ such that the automorphism fixes every element of $F$.
But, how do I show $|Aut(\mathbb{F}_{p^r}/\mathbb{F}_{p})|=r$? A simple, easy to grasp proof without holes in it would be ideal.

Every field $K$ of characteristic $p$ has the Frobenius endomorphism
$F:x\mapsto x^p$. This is a homomorphism of fields, and so is injective.
If $K$ is finite, then $F$ must be bijective, so an automorphism.
On $\Bbb F_p$, $F$ acts trivially.
The fixed points of $F^t$ are the solutions of $x^{p^t}-x$.
Every element of $\Bbb F_{p^r}$ is a solution of $x^{p^r}-x$,
if $t<r$ then not all elements of $\Bbb F_{p^r}$ is a solution of
$x^{p^t}-x$ since that polynomial has fewer than $p^r$ zeros.
Thus $F$ has order $r$ on $\Bbb F_{p^r}$. As $|\Bbb F_{p^r}:\Bbb F_p|
=r$, then the Galois group must consist of the powers of $F$.
