Question
Show that if $n$ is a positive integer greater than $1$, then the Mersenne number $M_n$ cannot be the power of a positive integer.
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Let $M_n = 2^n - 1$ be a Mersenne number. Suppose that $M_n = q^k$ for some $q \in \mathbb{N}$.
$M_n$ is odd $\implies q$ is odd.
Suppose that $k = 2k_1$. This implies that $M_n = 2^n - 1 = (q^{k_1})^2$.
But the square of an odd integer is congruent to $1$ modulo $4$. However, $2^n - 1$ is congruent to $3$ modulo $4$ for $n > 1$.
This is a contradiction. Therefore, $k$ has to be odd, and we have the equation $$2^n = a^k + 1 = (a + 1)(1 - a + a^2 - a^3 + \ldots + (-1)^{k-1} a^{k-1}) = (a + 1)(1 - a + a^2 - a^3 + \ldots + a^{k-1}).$$ This implies that we have $$a = 2^m - 1$$ for some $m$. This means that we have $$a^k = (2^m - 1)^k \implies 2^n - 1 = (2^m - 1)^k.$$
How should I proceed?