Theorem of Even Perfect Numbers/Sufficient Condition

Theorem
Let $n \in \N$ be such that $2^n - 1$ is prime.

Then $2^{n-1} \left({2^n - 1}\right)$ is perfect.

Proof
Suppose $2^n - 1$ is prime.

Let $a = 2^{n-1} \left({2^n - 1}\right)$.

Then $n \ge 2$ which means $2^{n-1}$ is even and hence so is $a = 2^{n-1} \left({2^n - 1}\right)$.

Note that $2^n - 1$ is odd.

Since all divisors (except $1$) of $2^{n-1}$ are even it follows that $2^{n-1}$ and $2^n - 1$ are coprime.

Let $\sigma \left({n}\right)$ be the sigma function of $n$, that is, the sum of all divisors of $n$ (including $n$).

From Sigma Function is Multiplicative, it follows that $\sigma \left({a}\right) = \sigma \left({2^{n-1}}\right)\sigma \left({2^n - 1}\right)$.

But as $2^n - 1$ is prime, $\sigma \left({2^n - 1}\right) = 2^n$ from Sigma of Prime Number.

Then we have that $\sigma \left({2^{n-1}}\right) = 2^n - 1$ from Sigma of Power of Prime.

Hence it follows that $\sigma \left({a}\right) = \left({2^n - 1}\right) 2^n = 2 a$.

Hence from the definition of perfect number it follows that $2^{n-1} \left({2^n - 1}\right)$ is perfect.