$p$ is a prime number.
I want to show $\alpha \in \mathbb{F}_{p^n}$ is in $\mathbb{F}_p$ if and only if $\alpha^p = \alpha$.
One way is essentially just Fermat's Little Theorem, but I'm confused on how to show the converse, that is, $\alpha^p = \alpha \implies \alpha \in \mathbb{F}_p$.
I know that
$$\mathbb{F}_{p^n} \cong \frac{\mathbb{F}_p}{(f)}$$
for $f \in \mathbb{F}_p[X]$ any irreducible polynomial of degree $n$.
I've tried playing around with writing $\alpha = a_{p-1} X^{p-1} + \dots + a_1 X + a_0$ for $a_0, \dots, a_{p-1} \in \mathbb{F}_p$, but to no avail. And if this does end up working somehow, it's certainly not very elegant. My understanding is that giving an explicit formula for $f$ is not really possible for general $p$, $n$. (https://en.wikipedia.org/wiki/Finite_field#Explicit_construction certainly suggests so.) So perhaps this approach won't work since we'd probably need to know something about $f$.
How could I approach this?