Field extensions and Galois goup

Question

Let $p$ b a prime, $n\in \mathbb{N}$ and let $f=x^{p^n}-x-1\in \mathbb{F}_p[x]$ be irreducible. Let $a\in \overline{\mathbb{F}}_p$ be a root pf $f$.

We have that $[\mathbb{F}_p(a):\mathbb{F}_p]=p^n$, where $\{1,a,\ldots , a^{p^n-1}\}$ is a basis of $\mathbb{F}_p(a)/\mathbb{F}_p$, so $\mathbb{F}_p(a)=\{c_0+c_1a+\ldots +c_{p^n-1}a^{p^n-1} : c_i\in \mathbb{F}_p\}$. There are $p$ choices for each $c_i$. So, there are $p\cdot p\cdot \ldots \cdot p=p^{p^n}$ choices for $(c_0, c_1, \ldots c_{p^n-1})$. Therefore, we have that $|\mathbb{F}_p(a)|=p^{p^n}$, right?

We have that $\mathbb{F}_p(a)$ contains all the roots of $f$.

We also have that for each $b\in \mathbb{F}_{p^n}$, $a+b$ is a root of $f$ and that $\mathbb{F}_{p^n}\leq \mathbb{F}_p(a)$ and $n=p^i$ for some $i\in \{0, 1, \ldots , n\}$.

Then I want to show that $\text{Gal}(\mathbb{F}_p(α)/\mathbb{F}_{p^n} )$ is cyclic and let $\tau$ be a generator. I want to calculate also the order of $\tau$ as a function of $i$.

We have that $|\mathbb{F}_{p^n}|=p^n$ and $|\mathbb{F}_p(a)|=p^{p^n}$. From the corollary

we have that $\mathbb{F}_p(α)/\mathbb{F}_{p^n} $ is Galois, $n\mid p^n$, which is true since $n=p^i$. From the corollary we also have that $\text{Gal}(\mathbb{F}_p(α)/\mathbb{F}_{p^n} )$ is cyclic and is generated by the automorphism $\tau (x)=x^{p^n}$.

Do we calculate the order of $\tau$ as follows?
$$\tau (x)=x^{p^n} \\ \tau^2 (x)=\tau( x^{p^n})=(\tau (x))^{p^n}=(x^{p^n})^{p^n}=x^{p^n\cdot p^n}=x^{p^{2n}} \\ \ldots \\ \tau^k(x)=x^{p^{kn}}$$

So that $m$ is the order, $m$ has to be the smallest integer such that $\tau^m(x)=x \Rightarrow x^{p^{mn}}=x$. Since $x\in \mathbb{F}_p(a)$ and $| \mathbb{F}_p(a)|=p^{p^n}$ we have that $x=x^{p^{p^n}}$. So, we get that $x^{p^{mn}}=x^{p^{p^n}} \Rightarrow p^{mn}=p^{p^n}\Rightarrow mn=p^n\Rightarrow m=\frac{p^n}{n}$.

Then independent from that I want to calculate a simple expression of $\tau^k(a)$.

I have done the following: Since $\tau$ is a generator of $\text{Gal}(\mathbb{F}_p(α)/\mathbb{F}_{p^n} )$ that means that $\tau$ is a $\mathbb{F}_{p^n}$-automorphism of $\mathbb{F}_p(a)$ that maps to a root of $f$ to an other root, right?

So, we have that $\tau (a)=a+b$. Therefore, we get the following: $$\tau^2(a)=\tau(a+b)=\tau (a)+\tau (b)=a+b+b=a+2b \\ \tau^3(a)=\tau(\tau^2(a))=\tau (a+2b)=\tau (a)+\tau (2b)=a+b+2b=a+3b \\ \ldots \\ \tau^k(a)=a+kb$$ Is this correct?

The order of $\tau$ is the smallest integer $m$ such that $\tau^m(a)=a\Rightarrow a+mb=a$. Does this imply that $m=p$ ?