Prove that if $2^n - 1$ is prime, then $n$ is prime for $n$ being a natural number
I've looked at https://math.stackexchange.com/a/19998
It is known that $2^n-1$ can only be prime if $n$ is prime. This is because if $jk=n$, $2^n-1=\sum_{i=0}^{n-1}2^i$ $=\sum_{i=0}^{j-1} 2^i \sum_{i=0}^{k-1} 2^{ij}$.
So they will only continue to alternate at twin primes. In particular, $2^{6k+2}-1, 2^{6k+3}-1$ and $2^{6k+4}-1$ will all be composite
What I dont understand is how do I get:
$$\sum_{i=0}^{n-1}2^i=\sum_{i=0}^{j-1} 2^i \sum_{i=0}^{k-1} 2^{ij}$$
If I am looking at the wrong answer, how can I proof the above?
Then again, maybe I am indeed looking at the wrong thing?
In particular, $2^{6k+2}-1, 2^{6k+3}-1$ and $2^{6k+4}-1$ will all be composite
This doesnt say why $2^{6k+2}-1$ is composite?