I am trying to show that if $n=2^{m-1}(2^{m}-1)$, where $m$ is a positive integer such that $2^{m}-1$ is composite, then $n$ is abundant. This is my proof thus far:
Proof. Let $n=2^{m-1}(2^{m}-1)$, and let $m$ be a positive integer such that $2^{m}-1$ is composite. It follows that $2^{m-1}$ is even and $2^{m}-1$ is odd. Therefore, $(2^{m-1},2^{m}-1)=1$, and $\sigma(n)=\sigma(2^{m-1}(2^{m}-1))=\sigma(2^{m-1})\sigma(2^{m}-1)=(2^{m}-1)\sigma(2^{m}-1)$.
However, I get stuck here. I know that all I am left to do is show that $\sigma(2^{m}-1)>2^{m}$, but I do not see how I could accomplish that (since $2^{m}-1$ is composite). What do you guys think?