Prove that, $\displaystyle {2^{(p - 1)}}({2^p} - 1)$ is perfect number if $\displaystyle ({2^p} - 1)$ is prime number
A number is perfect if the sum of it's divisors are twice the number.
Since by assumption
$\displaystyle 2^p-1$ is prime the has only two divisors 1 and itself.
Since the other factor is just powers of 2 its divisors are
$\displaystyle 1,2,4,...2^{p-1}$
Since the latter is a geometric series we get that
$\displaystyle \sum_{n=1}^{p}2^{n-1}=2^p-1$
Since the has already added the divisor 1 we just add other divisor
$\displaystyle (2^p-1)+(2^p-1)=2(2^p-1)$
Also, you should know that if $\displaystyle n$ is an even perfect number say $\displaystyle n=2^{k-1}m$ then $\displaystyle m=2^k-1$ and it is a prime number.
Proof:
$\displaystyle m\in \mathbb{N}$ odd, $\displaystyle k\geqslant 2 $.
$\displaystyle \text{gcd}(2^{k-1},m)=1$, hence:
$\displaystyle \sigma(n)=\sigma(2^{k-1}m)=\sigma(2^{k-1})\sigma(m)=(2^k-1)\sigma(m) $ (1)
$\displaystyle n$ is perfect number, therefor:
$\displaystyle \sigma(n)=2n=2^km$
With (1), we get:
$\displaystyle 2^km=(2^k-1)\sigma(m)$ (2)
From the above we can deduce that $\displaystyle m=(2^k-1)M$, now if we put this $\displaystyle m$ to (2) we will get:
$\displaystyle 2^k(2^k-1)M=(2^k-1)\sigma(m)$
Or:
$\displaystyle 2^kM=\sigma(m)$.
$\displaystyle m$ and $\displaystyle M$ are both divisors of m, and therefor:
$\displaystyle 2^kM=\sigma(m)\geqslant m+M=2^kM$.
The conclusion is that: $\displaystyle \sigma(m)=m+M$.
From the above follows that $\displaystyle m$ is prime number and $\displaystyle M=1$
Or: $\displaystyle m=(2^k-1)M=2^k-1$