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 $a^kb^k=(ab)(a^{k1}+...+b^{k1})$ – mrprottolo Jan 03 '16 at 11:18
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If $p = a\cdot b $ where $a \neq 1, b \neq 1$ then $n = 2^{ab}  1 = (2^a)^b  1 = (2^a1)(2^{a(b1)} + ...+ 1)$ 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 $2^k12^p1$ and $2^k1 > 1$, so $2^p1$ cannot be a prime as desired.
Gyumin Roh
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If $1<x\le y$ then $2^x1$ divides $2^{xy}1$. Moreover $1<2^x1<2^{xy1}$, which is enough to conclude.
Paolo Leonetti
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