If $p^k-1$ divides $p^n-1$, then $\mathbb F_{p^k}\subset \mathbb F_{p^n}$

Question

Suppose $p$ is a prime and $p^k-1$ divides $p^n-1$. How can I show that $\mathbb F_{p^k}\subset \mathbb F_{p^n}$? The only thing I can deduce from the given condition is that $\mathbb F_{p^n}^\times$ contains an element of order $p^k-1$ since $\mathbb F_{p^n}^\times$ is cyclic of order $p^n-1$, and any cyclic group $C_r$ contains an element of order $s$ provided $s$ divides $r$ (take $x^{r/s}$ where $x$ is a generator). How can I proceed?

Answer 1

Hint: $\mathbb F_{p^n}$ is the set of roots of $x^{p^n}-x$.

Answer 2

You can show by induction on $\max(k,n)$ that $\gcd(p^k-1, p^n-1) = p^{\gcd(k,n)}-1$. In particular, $p^k-1\mid p^n-1$ iff $k\mid n$. It is well-known that $\mathbb{F}_{p^k}\leq \mathbb{F}_{p^n}$ iff $k\mid n$.