Prove that if k and n are positive integers with n>k>1 such that k divides n, then ((2^k)-1) divides ((2^n)-1)
In general if $\displaystyle k|n$ then $\displaystyle a^k-1$ divides $\displaystyle a^n-1$.
This is because $\displaystyle n=km$ thus $\displaystyle a^{km}-1 = (a^k)^m - 1$.
Now use identity for factoring $\displaystyle x^n - y^n$.