Can someone give me a short proof of the identity, $$\sum_{n=1}^\infty\frac{q^nx^{n^2}}{1qx^{n}}+\sum_{n=1}^\infty\frac{q^nx^{n(n+1)}}{1x^n}=\sum_{n=1}^\infty\frac{q^nx^n}{1x^n}$$
It's just about using geometric series a lot. Indeed, we have $\sum\limits_{n=1}^\infty\frac{q^n x^{n^2}}{1qx^n}=\sum\limits_{n=1}^\infty\sum\limits_{k=0}^\infty q^{n+k}x^{n(n+k)}=\sum\limits_{m=1}^\infty\sum\limits_{d\ge m}q^dx^{md}$ and $\sum\limits_{n=1}^\infty\frac{q^n x^{n(n+1)}}{1x^n}=\sum\limits_{n=1}^\infty\sum\limits_{k=0}^\infty q^{n}x^{n(n+1+k)}=\sum\limits_{m=1}^\infty\sum\limits_{1\le d<m}q^dx^{md}$. Therefore, $\sum\limits_{n=1}^\infty\frac{q^n x^{n^2}}{1qx^n}+\sum\limits_{n=1}^\infty\frac{q^n x^{n(n+1)}}{1x^n}= \sum\limits_{m=1}^\infty\sum\limits_{d\ge m}q^dx^{md}+\sum\limits_{m=1}^\infty\sum\limits_{1\le d<m}q^dx^{md}= \sum\limits_{m=1}^\infty\sum\limits_{d=1}^\infty q^dx^{md}$. However, $\sum\limits_{n=1}^\infty\frac{q^n x^{n}}{1x^n}= \sum\limits_{m=1}^\infty\sum\limits_{d=1}^\infty q^dx^{md}$ as well, and we are done. 

