In particular, for $prime$ $n$, does there exist a bound on $n$ beyond which $(2^n-1^n)$ will always have at least one more factor apart from trivial factor $(2^1-1^1)$?
Factoring $(2^n-1^n)$ as $n$ approaches infinity
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algebra-precalculus
number-theory
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1Well, $2^1-1^1=1$ is not *really* a factor. That being said, [it is an open problem](https://en.wikipedia.org/wiki/Mersenne_prime) whether $2^p-1$ is prime for infinitely many primes $p$. – 2017-02-04
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0What is the point of the exponent on the 1? – 2017-02-04
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0Related: look up Mersenne Prime on wikipedia. – 2017-02-04