[nooonuii]

To make life easier, I conjecture that

$\displaystyle{2^{2^n-1}-2^n-1}$ is divisible by $p$

where $p$ is the smallest prime divisor of $2^n-1$.

I can prove two special cases for this conjecture :

Case 1 : n is even. Then $\displaystyle{2^{2^n-1}-2^n-1}$ is divisible by $3$.
Case 2 : n is odd and $2^n-1$ is prime, i.e. a Mersenne prime. Then $\displaystyle{2^{2^n-1}-2^n-1}$ is divisible by $2^n-1$ by Fermat's Little Theorem.

Thus it remains to prove the case where n is odd and $2^n-1$ is composite.