Show that $$\begin{align} &\prod\limits_{k\ge 1}\frac{1-q^k}{1+q^k}=\sum \limits_{n\in \mathbb{Z}}(-1)^{n}q^{n^2}\tag{1}\\ &\prod\limits_{k\ge 1}(1-q^k)^3=\sum \limits_{n\ge 0}(-1)^{n}(2n+1)q^{n(n+1)/2}\tag{2}\end{align}$$
I have no idea of how to prove this two beautiful identity? I guess this two identity need help with Jacobi triple product?
The first one is given in Prove these identities using Jacobi's triple product identity.
As regards the second one, use again the Jacoby triple product in the form $$\prod_{n\geq1}(1-x^{2n})(1-yx^{2n+1})(1-y^{-1}x^{2n-1}) =\frac{1}{1-yx}\sum_{n=-\infty}^{\infty}(-y)^nx^{n^{2}},$$ and note that, $$\begin{align} \lim_{y\to x^{-1}}\left(\frac{1}{1-yx}\sum_{n=-\infty}^{\infty}(-y)^nx^{n^{2}}\right)&\stackrel{H}{=} \lim_{y\to x^{-1}}\left(\frac{1}{-x}\sum_{n=-\infty}^{\infty}n(-y)^{n-1}(-1)x^{n^{2}}\right)\\ &=\sum_{n=-\infty}^{\infty}(-n)(-1)^{n}x^{n^{2}-n}\\ &=\sum_{n=1}^{\infty}n(-1)^{n}x^{n^{2}+n}+\sum_{n=1}^{\infty}n(-1)^{n-1}x^{n^{2}-n}\\ &=\sum_{n=0}^{\infty}n(-1)^{n}x^{n^{2}+n}+\sum_{n=0}^{\infty}(n+1)(-1)^{n}x^{(n+1)^{2}-(n+1)}\\ &=\sum_{n\geq 0}(-1)^n(2n+1)x^{n^{2}+n}. \end{align}$$ Hence, by letting $y\to x^{-1}$ in the triple product, we get $$\prod_{n\geq1}(1-x^{2n})(1-x^{2n})(1-x^{2n}) =\sum_{n\geq 0}(-1)^n(2n+1)x^{n^{2}+n}.$$ Finally replace $x$ with $q^{1/2}$.