Sequence: $a_n=2^n-1$
I want to prove that $a_m|a_n$ whenever $m|n$.
I started by approaching with induction, base/trivial cases being $m=n$, $n=0$, and $m=1$, but I'm not sure where to go from there or if there is a more straightforward method.
Note:
$m|n$ means for some $x$, $mx = n$
$a_m|a_n$ means for some $y$, $a_my=a_n$
Other similar questions appeal to laws beyond basic arithmetic and the definitions of divisibility and the sequence.
If $m | n$ then $n = km$ so
$\begin{array}\\ 2^n-1 &=2^{km}-1\\ &=(2^{m})^k-1\\ &=(2^m-1)\sum_{j=0}^{k-1}2^{jm}\\ \end{array} $