How could I prove that $(-q;q^2)_\infty (q;q)_\infty = 1 + 2 \sum_{i=1}^\infty (-1)^i q^{2 i^2}?$ If that is too difficult is there a way to show $(-q;q^2)_\infty (q;q)_\infty \equiv 1 \pmod 2?$
edit The identity $(-q;q^2)_\infty (q;q)_\infty = (q^2;q^2)_\infty (q^2;q^4)_\infty$ is easily found and lets you substitute $\rho = q^2$ to reduce the problem to $(\rho;\rho)_\infty (\rho;\rho^2)_\infty = \sum_{i = -\infty}^{\infty} (-1)^{i} \rho^{i^2}.$
Definition: $(a;q)_n = \prod_{k=0}^n \left(1 - aq^k\right)$