Can someone please verify whether my proof is logically correct? :)
Let $p\geq 2$. If $2^{p}-1$ is prime, then $p$ must also be prime.
Proof:
Let $p=a\cdot b$ with $a>1$ and $b>1$ being integers. Then
$2^{p}-1 =2^{ab}-1=(2^{a}-1)(2^{b(a-1)}+2^{b(a-2)}+\cdots +2^{a}+1) = (2^{b}-1)(2^{a(b-1)}+2^{a(b-2)}+\cdots +2^{b}+1)$
where $2^{a}-1>1$ and $2^{b}-1>1$ since $a>1$ and $b>1$. If p is composite with $p=a\cdot b$ and $1<a<p$, then $2^{p}-1$ is composite, since $2^{ab}-1$ is divisible by $2^{a}-1$. $\square$
I don't understand why it is showing if $p$ is composite, then $2^{p}-1$ is composite though.