Prove that if $m$ divide $n$ then $x^{p^{m-1}}-1$ divide $x^{p^{n-1}}-1$

Question

I have been trying for several days to prove the equality of the title, but without any positive results. I know I have to first prove that if m divides n (it has to be multiple of m) then $p^{m}-1$ divide $p^{n}-1$.

That is the one that I have been able to demonstrate knowing that there is a quotient for this division that has zero rest and is the following: $\sum_{i=1}^{\alpha}(p^{(\alpha-i)\times m})$.

Someone could help me relate it to the title and get the proof.

Thanks.

Answer 1

Write $n=mk$ for $k\in\mathbb{N}$. Now we have $$p^n-1=(p^m)^k-1=(p^m-1)\sum_{j=0}^{k-1}(p^m)^j$$ This is an application of the factorization identity $$(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+\cdots+1)$$ which you should definitely memorize.