Let $n=(2^{p-1})\cdot(2^p-1)$ where n is an even prefect number and p is prime.
I know the divisors are $1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1$, and $n$.
I get $(2^{(1/2)p(p-1)}\cdot ((2^p)-1)\cdot(2^{p-1} \cdot(2^p-1))$ but I can't turn it into $n^p$.
Any help would be appreciated.
Let $n = 2^{p-1} (2^p - 1)$ denote an even perfect number.
Let $T(n)$ denote the product of divisors of $n$.
Then, in general, we have the formula $$T(n) = n^{d(n)/2}$$ where $d(n)$ is the number of divisors of $n$.
Let us use the formula for the number of divisors of $n$, $d(n)$:
Since $n = 2^{p-1} (2^p - 1)$, then $$d(n) = (1+1)((p - 1) + 1) = 2p.$$
Therefore, the product of divisors of $n$, is given by: $$T(n) = n^{d(n)/2} = n^{2p/2} = n^p.$$
Reference:
On the product of divisors of a positive integer - Tibor Šalát and Jana Tomanová, Mathematica Slovaca (2002) Volume: 52, Issue: 3, page 271-287