Let $p$ be a prime, $n\in \mathbb{N}$ and $f=x^{p^n}-x-1\in \mathbb{F}_p[x]$ irreducible.
Let $a\in \overline{\mathbb{F}}_p$ (algebraic closure of $\mathbb{F}_p$) be a root of $f$.
We have that $\mathbb{F}_p(a)$ contains all the roots of $f$, for each $b\in \mathbb{F}_{p^n}$, $a+b$ is a root of $f$ and $\mathbb{F}_{p^n}\subseteq \mathbb{F}_p(a)$.
How can we show that $n=p^i$ for any $i\in \{0, 1, \ldots , n\}$?
I want to show that $Gal(\mathbb{F}_p(a)/\mathbb{F}_{p^n})$ is cyclic and let $\tau$ be a generator.
Since $a+b$ is a root of $f$ for each $b\in \mathbb{F}_{p^n}$, we have that $f$ is separable.
So, $\mathbb{F}_p(a)$ is the splitting field of the separable polynomial $f\in \mathbb{F}_p[x]$.
Therefore the extension $\mathbb{F}_p(a)/\mathbb{F}_p$ is Galois, with $|Gal(\mathbb{F}_p(a)/\mathbb{F}_p)|=[\mathbb{F}_p(a):\mathbb{F}_p]=\deg f=p^n$.
How could we continue?
We know from your other topic that this is a Galois extension. Now consider $|K|=|\Bbb F_p(a)| = p^k(=p^{p^n})$, then $[K:\Bbb F_p]=k \; (=p^n)$. We also know that the defining characteristic of elements of $K$ is that $x^{p^k}-x=0$, and $k$ is the smallest positive integer for which this holds for all $\alpha\in K$. But then the field automorphism $x\mapsto x^p$ has order $k$ in the Galois group. But as $|\text{Gal}(K/\Bbb F_p)|=k$, we see that this automorphism generates the group.
Note: Just as in your previous topic, there is absolutely nothing special about the polynomial you chose: this is true for all finite extensions of finite fields.