Prove that if $d$ divides $n$, then $2^d -1$ divides $2^n -1$.
Use the identity $x^k -1 = (x-1)*(x^{k-1} + x^{k-2} + \cdots + x +1)$
Prove that if $d$ divides $n$, then $2^d -1$ divides $2^n -1$.
Use the identity $x^k -1 = (x-1)*(x^{k-1} + x^{k-2} + \cdots + x +1)$