Prove that if $\mathbb{F}_{p^n} \subseteq \mathbb{F}_{p^m}$ then $n \mid m$

Question

The order of $\mathbb{F}_{p^n}$ is $p^n$ and $\mathbb{F}_{p^m}$ is $p^m$. Since $\mathbb{F}_{p^m}$ is the largest field and it contains all the elements of $\mathbb{F}_{p^n}$ and the order of any element of $\mathbb{F}_{p^n}$ divides also the order of $\mathbb{F}_{p^m}$. How I proceed further?

Answer 1

This is a straightforward application of the multiplicativity of the degree: If $\mathbb{F}_{p^n} \subseteq \mathbb{F}_{p^m}$ then $$ m = [\mathbb{F}_{p^m} : \mathbb{F}_p] = [\mathbb{F}_{p^m} : \mathbb{F}_{p^n}] \cdot [\mathbb{F}_{p^n} : \mathbb{F}_p] = [\mathbb{F}_{p^m} : \mathbb{F}_{p^n}] \cdot n, $$ so $n \mid m$.