A number $n\in N$ Show that if $p$ is a prime, such that $2^p - 1$ is also a prime, a Mersenne prime that is, then $n = 2^{p-1}(2^p-1)$ is a perfect number.
So I know that $n$ must be divisible by, $2^p-1$, $2^{p-1}$ and $1$ (the proper divisors of $n$), which sum up to be $2^p-1 + 2^{p-1} + 1 = 2^p + 2^{p-1}$. I'm stumped on where to go from here, since the sum isn't equal to $n$.
Assuming $2^p-1$ is prime number then $2^{p-1} (2^p -1)$ is perfect number if
$\sigma(2^{p-1} (2^p -1)) = \sigma(2^{p-1}) \sigma(2^p-1) = 2^p \sigma(2^{p-1}) = 2^p (2^{p}-1) = 2 * 2^{p-1} (2^{p}-1) $ twice the number we start with so its perfect number.
Done.