I am just trying to solve the first part
I think I understand how to attempt this problem using a specific p, k, and n, my steps are as follows:
We know that the field $\mathbb F_{p^m}$ = the roots of $x^{p^m} - x$ and for any $a \in \mathbb F_{p^m}$, $a^{p^m} - a = 0$
Using p=2, k=2, and n=4,
We must prove that $x^4 - x \subseteq x^{16} - x$
By dividing $x^4 - x$ into $x^{16} - x$ we get the quotient $(x^{12} + x^9 + x^6 + x^3 + 1) = q(x)$ with no remainder.
Then, we can rewrite
$f(x) = (x^{16} - x) = (x^4 - x)(q(x))$
Substituting for a, where we say for any $a \in \mathbb F_{p^m}, a^{p^m} - a = 0,$ so $(a^4 - a) = 0.$
$f(a) = (a^{16} - a) = (a^4 - a)(q(a)) = 0$
so $a$ is a root of $\mathbb F_4$ and $\mathbb F_{16}$ which means $\mathbb F_4 \subseteq \mathbb F_{16}$
At this point I am unsure of to translate this to an arbitrary p prime, k, and n. Any guidance would be greatly appreciated!
$k|n \Rightarrow p^k-1 | p^n-1 \Rightarrow x^{p^k-1}-1 | x^{p^n-1}-1\Rightarrow x^{p^k}-x | x^{p^n}-x$.
Edit
Since $x^{p^k}-x | x^{p^n}-x$. Then $$x^{p^n}-x=(x^{p^k}-x)q(x)$$
Therefore $(a^{p^k}-a)=0 \Rightarrow (a^{p^n}-a)=0$