I'm trying to follow a solution I'm reading. The idea is to prove $x^{p^n}-x+1$ is irreducible over $\mathbb{F}_p$ when $n=1$ or $n=p=2$. The solution is as follows:
If $x^{p^n}-x+1$ is irreducible, and $\alpha$ a root, $\mathbb{F}_p(\alpha)\simeq \mathbb{F}_{p^{p^n}}$. This is a splitting field for $x^{p^{p^n}}-x$, and since $x^{p^n}-x+1$ is a factor, $\mathbb{F}_p(\alpha)$ contains all the roots of $x^{p^n}-x+1$. Also, $\alpha+a$ is a root for all $a\in \mathbb{F}_{p^n}$, so $\mathbb{F}_{p^n}\subset\mathbb{F}_p(\alpha)$. And $[\mathbb{F}_p(\alpha):\mathbb{F}_{p^n}]=p^{n-i}$, where $n=p^i$, and is a cyclic extension. There is a subgroup of $\operatorname{Gal}(\mathbb{F}_p(\alpha)/\mathbb{F}_{p^n})$ isomorphic to $\mathbb{Z}/p\mathbb{Z}$, generated by a root $\beta$ of $x^{p^n}-x+1$. Then $[\mathbb{F}_p(\beta):\mathbb{F}_{p^n}]=p$.
The solution can be found and the very end of this document.
I don't follow what they mean when they say the subgroup isomorphic to $\mathbb{Z}/p\mathbb{Z}$ is generated by a root $\beta$, as $\beta$ is an element of the field, and the subgroup is a group of automorphisms. And why does that imply $[\mathbb{F}_p(\beta):\mathbb{F}_{p^n}]=p$? My understanding is that the degree of the field extension of $\mathbb{F}_{p^n}$ should be equal to the index of the subgroup, which would be $p^{n-i-1}$?