Notice $2^{n-1}-1=(2^k)^p-1=(2^k-1)\cdot\Phi_p(2^k)$. Let $q $ be a prime divisor of $n $ obviously $q\neq p $ and $q\nmid k $. So assume $q\mid \Phi_p(2^k) $ so it follows $ord_q (2^k)=p $ i.e we have $p\mid q-1$ for any prime divisor of $n $. So this implies $p\nmid k $ otherwise we have $q\mid 2^k-1$ and we are done. So $\Phi_p (2^k)=\prod_{d\mid k} \Phi_{pd}(2)=\prod_{d\mid k} \frac {\Phi_d (2^p)}{\Phi_d (2)} $ so this contradicts the information for the $ord_q $ so $n\mid 2^k-1$ i.e $2^k>2^k-1\ge n $.