Let's consider two polynomials in $\mathbb{F}_{p}[x]$: $f(x)=x^{p^{n}-1}-1$ and $g(x)=x^{p^{k}-1}-1$. How to prove that $g(x) \mid f(x)$ iff $k \mid n$?
For instance, it's worth trying this: According to the Little Fermat theorem $x^{p} \equiv x \mod p$, then $x^{p^{k}} \equiv x^{k}\mod p$, $ x^{p^{k}-1} \equiv x^{k-1} \mod p$. So the problem is reduced to showing that $x^{k-1}-1$ divides $x^{n-1}-1$ iff $k | n$.
Any help would be much appreciated.
If $k\mid n$, then every root of $x^{p^k-1}-1$ is also a root of $x^{p^n-1}-1$, so the field $\mathbb{F}_{p^k}$ is a subfield of $\mathbb{F}_{p^n}$, by recalling that, in an algebraic closure of $\mathbb{F}_p$, the field with $p^r$ elements is uniquely determined as the set of roots of $x^{p^r}-x$.
Let's see the converse. If $\mathbb{F}_{p^k}$ is a subfield of $\mathbb{F}_{p^n}$ then the group of invertible elements of the former is a subgroup of the group of invertible elements of the latter, which implies $$ p^k-1\mid p^n-1 $$ This in turn implies that $k\mid n$. Let $n=kq+r$, with $0\le r<k$; then $$ p^n-1=(p^k)^qp^r-1=(p^k-1+1)^qp^r-1= -1+p^r\sum_{i=0}^q\binom{q}{i}(p^k-1)^i $$ so we conclude that $p^k-1$ divides $p^r-1$ which is possible only if $r=0$.
One way to do this is: $$n|m\implies X^n-1|X^m-1 \tag 1$$ to prove it take $m=nt$ then : $$X^m-1=(\color{#0a0}{X^n})^t-1=(\color{#0a0}{X^n}-1)\left(\sum_{i=0}^{t-1}(\color{#0a0}{X^n})^i\right) \tag 2$$ Or you can see this as a consequence of $Y-1|Y^t-1$ for $Y=X^n$. Another result which follows from this is that if you replace $X$ by a specific number in $(1)$ the implication still holds, so we can conclude about your final question in two steps:
$$k|n\overset{\ \large \color{#c00}{\rm (1) \text{ for }X=p}}\implies p^k-1|p^n-1\overset{\ \large \color{#c00}{\rm (1) \text{ for } X=x}}\implies x^{p^k-1}-1|x^{p^n-1}-1 \tag 3$$
