My first thought was to try a contradiction; So given n is prime assume p is not prime i.e $p = p_{1}^{\alpha1} .... p_{r}^{\alpha r}$. But i didnt know where to go from there.
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Try to use that $a^k-b^k=(a-b)(a^{k-1}+...+b^{k-1})$ – mrprottolo Jan 03 '16 at 11:18
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If $p = a\cdot b $ where $a \neq 1, b \neq 1$ then $n = 2^{ab} - 1 = (2^a)^b - 1 = (2^a-1)(2^{a(b-1)} + ...+ 1)$ is clearly composite.
DeepSea
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Assume that $p$ is not a prime, and let $k$ be $p$'s smallest prime divisor.
We have $2^k-1|2^p-1$ and $2^k-1 > 1$, so $2^p-1$ cannot be a prime as desired.
rkm0959
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If $1<x\le y$ then $2^x-1$ divides $2^{xy}-1$. Moreover $1<2^x-1<2^{xy-1}$, which is enough to conclude.
Paolo Leonetti
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