[step: The Frobenius map is a field automorphism of $\mathbb{F}_{p^n}$] Define $\operatorname{Fr} \colon \mathbb{F}_{p^n} \to \mathbb{F}_{p^n}$ by $\operatorname{Fr}(x) = x^p$. We verify it is a field homomorphism. Clearly $\operatorname{Fr}(1) = 1$ and $\operatorname{Fr}(xy) = (xy)^p = x^p y^p = \operatorname{Fr}(x)\operatorname{Fr}(y)$. For addition, the Frobenius identity in characteristic $p$ gives \begin{align*} \operatorname{Fr}(x + y) = (x + y)^p = x^p + y^p = \operatorname{Fr}(x) + \operatorname{Fr}(y), \end{align*} since every binomial coefficient $\binom{p}{k}$ with $0 < k < p$ is divisible by $p$ and therefore vanishes in $\mathbb{F}_{p^n}$. As a nonzero field homomorphism, $\operatorname{Fr}$ is injective. Because $\mathbb{F}_{p^n}$ is finite, an injective map from a finite set to itself is surjective, so $\operatorname{Fr}$ is a bijection and hence a field automorphism. [guided: Why does Fr fix F_p?] For any $a \in \mathbb{F}_p$ we have $a^p = a$ by Fermat's little theorem, so $\operatorname{Fr}(a) = a$. Thus $\operatorname{Fr}$ fixes $\mathbb{F}_p$ pointwise and belongs to $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)$. [step: The Frobenius has order exactly $n$ in the Galois group] We show $\operatorname{ord}(\operatorname{Fr}) = n$. The $k$-th iterate of the Frobenius acts by $\operatorname{Fr}^k(x) = x^{p^k}$. Now $\operatorname{Fr}^k = \operatorname{id}$ if and only if $x^{p^k} = x$ for every $x \in \mathbb{F}_{p^n}$, which happens if and only if every element of $\mathbb{F}_{p^n}$ is a root of $t^{p^k} - t$. The elements of $\mathbb{F}_{p^n}$ are precisely the roots of $t^{p^n} - t$, so this condition holds if and only if $t^{p^n} - t$ divides $t^{p^k} - t$ in $\mathbb{F}_p[t]$, which is equivalent to \begin{align*} n \mid k. \end{align*} The smallest positive $k$ satisfying $n \mid k$ is $k = n$, so $\operatorname{ord}(\operatorname{Fr}) = n$. [guided: Why does $t^{p^n} - t \mid t^{p^k} - t$ iff $n \mid k$?] Every root of $t^{p^n} - t$ is a root of $t^{p^k} - t$ if and only if $\mathbb{F}_{p^n} \subseteq \mathbb{F}_{p^k}$, which holds if and only if $n \mid k$. This is the standard subfield criterion for finite fields. [step: The Galois group is cyclic of order $n$, generated by Frobenius] Since $\mathbb{F}_{p^n}/\mathbb{F}_p$ is a splitting field of $t^{p^n} - t$ over $\mathbb{F}_p$, the extension is Galois with \begin{align*} \lvert \operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) \rvert = [\mathbb{F}_{p^n} : \mathbb{F}_p] = n. \end{align*} We have produced an element $\operatorname{Fr} \in \operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)$ of order $n$. A group of order $n$ containing an element of order $n$ is cyclic, generated by that element. Therefore \begin{align*} \operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) = \langle \operatorname{Fr} \rangle \cong \mathbb{Z}/n\mathbb{Z}. \qquad \blacksquare \end{align*}