Euclid’s theorem about perfect numbers* depends on the prime divisor property, which will be proved in the next exercise.
Assuming this for the moment, it follows that if $\displaystyle 2^n-1$ is a prime $\displaystyle p$,
then the proper divisors of $\displaystyle 2^{n-1}p$ are $\displaystyle 1,2,2^2,...,2^{n-1}$ and $\displaystyle p,2p,2^2p...,2^{n-2}p$.

Given that the divisors of $\displaystyle 2^{n-1}p$ are those just listed, show that $\displaystyle 2^{n-1}p$ is perfect when $\displaystyle p=2^n-1$ is prime.
I've done that (just summing all the listed proper divisors and checking $\displaystyle \sum divisors=2^{n-1}p$).

That if $\displaystyle 2^n-1$ is prime, then
$\displaystyle 2^{n-1}(2^n-1)$ is perfect.

We can now fill the gap in the proof of Euclid’s theorem on perfect numbers* (above exercise), using the prime divisor property.

Use the prime divisor property* to show that the proper divisors of $\displaystyle 2^{n-1}p$, for any odd prime $\displaystyle p$, are $\displaystyle 1,2,2^2,...,2^{n-1}$ and $\displaystyle p,2p,2^2p...,2^{n-2}p$.
Help with this one.

* If $\displaystyle p$ is a prime that divides $\displaystyle ab$, then $\displaystyle p$ divides $\displaystyle a$ or $\displaystyle b$.