Subfields of the finite field: Is divisibility of multiplication groups a sufficient condition?

Question

For instance, $\mathbb{Z}_2^3$ does not contain a (copy of) $\mathbb{Z}_2^2$, because of divisibility: multiplication groups have $ 7 $, resp. $3$ elements and $ 3\nmid 7 $

But is true that if $(p^k-1) | (p^n -1) $ ($ p $ prime, $ n>k $ integer), then $\mathbb{Z}_p^n$ contains a copy of $\mathbb{Z}_p^k$ ?

Answer 1

The $p^k$ roots of $X^{p^k}-X$ form a subfield in that case.