Prove $\prod_{n \in \Bbb{N} } \left(\frac{1-q^n}{1+q^n}\right)^{(-1)^n} = \sum_{n \in \Bbb{Z}} q^{n^2}$ Let $q$ be a complex number with $|q| < 1$, prove that
$$
\prod_{n \in \Bbb{N} } \left(\dfrac{1-q^n}{1+q^n}\right)^{(-1)^n} = \sum_{n \in \Bbb{Z}} q^{n^2}
$$
Not sure if this helps but the LHS can also be written as 
$$
\prod_{n\in \Bbb{N}\\\text{n odd}}(1+q^n)(1+q^n+q^{2n}+\cdots)\prod_{n \in \Bbb{N} \\\text{n even}} (1-q^n)(1-q^n+q^{2n}-\cdots)
$$
I've tried the method here and the method here, but they both give me nightmares. Is there any good idea?

Note that
$$\begin{align}\prod_{n \in \Bbb{N} } \left(\dfrac{1-q^n}{1+q^n}\right)^{(-1)^n}&=
\prod_{n \in \Bbb{N} } \left(\dfrac{1-q^{2n}}{1+q^{2n}}\right)\left(\dfrac{1+q^{2n-1}}{1-q^{2n-1}}\right)\\
&=\prod_{n \in \Bbb{N} } \dfrac{1+q^n}{1+q^{2n}}\dfrac{1-q^{n}}{1-q^{2n-1}}(1+q^{2n-1})\\
&=\prod_{n \in \Bbb{N} } \dfrac{1+q^n}{1+q^{2n}}\prod_{n \in \Bbb{N} }\dfrac{1-q^{n}}{1-q^{2n-1}}\prod_{n \in \Bbb{N} }(1+q^{2n-1})\\
&=\prod_{n \in \Bbb{N} }(1+q^{2n-1})\prod_{n \in \Bbb{N} }(1-q^{2n})\prod_{n \in \Bbb{N} }(1+q^{2n-1})\\
&=
\prod_{n \in \Bbb{N} }(1-q^{2n})(1+q^{2n-1})^2\\
&=\sum_{n \in \Bbb{Z}} q^{n^2}
\end{align}$$
where at the last step we applied the Jacobi triple product for $x=q$ and $y=1$.
