For $d|n$, $x^{p^d} =x$ implies $x^{p^n} = x$.

Question

On the proving following theorem about finite fields

$\mathbb{F}_{p^n} \supset \mathbb{F}_{p^d}$ iff $d|n$

I have some problems. I know $\Rightarrow$ is trivial by considering extension degree.

But I have a problem on $\Leftarrow$ direction.

The textbook says

Suppose $d|n$ then if $x^{p^d} =x$ then also $x^{p^n} =x$, thus $\mathbb{F}_{p^d} \subset \mathbb{F}_{p^n}$.

But i am not sure why $x^{p^d} = x$ iplies $x^{p^n} =x$.

For example $d|n$ implies $n=dk$ for some $k$, then $x^{p^d} = x^{p^{nk}} = x$ then why this implies $x^{p^n}=x$?

Answer 1

We have $$x^{p^{2d}}=(x^{p^d})^{p^d}=(x)^{p^d}=x.$$ Now you can use induction to get your claim.

Answer 2

Induction using $$ x^{p^{kd}}=x^{p^dp^{(k-1)d}}=(x^{p^{d}})^{p^{(k-1)d}}=x^{p^{(k-1)d}}$$