$2^r-1$ divides $2^{rs}-1$

Question

For a composite number $n, n=rs,$ where $n>s≥r≥1.$ Show that $2^r-1$ divides $2^n-1$.

Thank you.

Answer 1

Thanks to this website for it inspired me: https://zhidao.baidu.com/question/327298334.html and GReyes' comment.

$2^{rs}-1=(2^r-1)(2^{r(s-1)}+2^{r(s-2)}+2^{r(s-3)}+\dots+2+1)$

This certainly divides $2^r-1$

Answer 2

Simply use $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})$