Source: Somewhere in the DLMF the below was quoted without proof or reference (I forgot the exact location sorry)
Prove that for $|q|<1$, $$\sum\limits_{m=1}^\infty \frac{q^{m(m+1)/2}}{(1-q)(1-q^2)\cdots (1-q^m)}=\sum\limits_{m=1}^\infty q^m(1+q)\cdots(1+q^{m-1}).$$
Attempt: It is easy to see, denoting $S(n,m)$ as Stirling numbers of the second kind, that
$$\sum\limits_{n=0}^\infty S(n,m)q^n=\frac{1}{(1-q)\cdots(1-q^m)},$$
so
$$\sum\limits_{m=1}^\infty \frac{q^{m(m+1)/2}}{(1-q)(1-q^2)\cdots (1-q^m)}=\sum\limits_{m=1}^\infty \sum\limits_{n=0}^\infty S(n,m)q^nq^{m(m+1)/2}.$$
Moreover, we have
$$\prod\limits_{i=1}^\infty (1+q^i) = \prod\limits_{i=1}^\infty \frac{1-q^{2i}}{1-q^i}=\prod\limits_{\text{odd } i} \frac{1}{1-q^i}=\sum\limits_{m=0}^\infty \sum\limits_{n=0}^\infty S(n,m)q^n$$
which maybe helps us manipulate the RHS but I got stumped here.
Question: Any help for proving the identity would be very much appreciated! Bonus points if it is somewhat in the neighborhood of my very imcomplete thought process.