Let $f\in\mathbb{F}_p(x)$ be a monic irreducible polynomial, denoting $\deg(f)=n$.
I wish to show (if it's true) that $f(x)$'s splitting field is $\mathbb{F}_{p^n}$.
I did some manual test for some polynomials till degree $5$, using the expansion field $\mathbb{F}_p[x]/\langle f(x) \rangle$ , and used Frobenius' automorphism to show that if $a$ is a root then $a^p , a^{p^2},a^{p^3}...a^{p^n}$ are unique roots. When $|\mathbb{F}_p[x]/\langle f(x) \rangle| = p^n$. Is there a way to show that this generally?
Other explanations will be appreciated, however I am not familiar with Galois theory.
All right, after some reading as proposed in comments, I believe this is the answer, please let me know If I'm wrong:
let $a$ be a root of $f(x)$ in $\mathbb{F}_p[x]/\langle f \rangle$ .
So $|\mathbb{F}_p[x]/\langle f \rangle| = p^n$. Thus it is isomorphic to $\mathbb{F}_{p^n}$ in which $x^{p^n}-x$ is splitting.
Thus $a$ is a root of $x^{p^n}-x\in \mathbb{F}_p[x]$.
$f$ is irreducible thus $f(x) | p(x)$ over $\mathbb{F}_p[x]$ because $f$ divides any polynomial with $a$ as a root. $x^{p^n}-x$ is splitting in $\mathbb{F}_p[x]/\langle f \rangle\cong\mathbb{F}_{p^n}$ thus $f$ must split in $\mathbb{F}_p[x]/\langle f \rangle$.
There is not subfield in which $f$ splits in, because any field containing a roots of $f$ , $\mathbb{F}_p(a)$ degree is $n$ and thus include $p^n$ elements.