Prove that $\nu(n) \le \nu(2^{n}-1)$ where $\nu(n)$ is the number of positive divisors of n

Question

Prove that $\nu(n) \le \nu(2^{n}-1)$

Ok so I have very few ideas for this question. I thought if I could find the prime decomposition of $2^{n}-1$, I could go somewhere with the formula for $\nu$, but I've had little success. Any help would be much appreciated.

Answer 1

Prove that if $d$ divides $n$, then $2^d-1$ divides $2^n-1$.