Prove these identities using Jacobi's triple product identity. I am requesting help with deriving some identities from Jacobi's triple product identity:
$$\sum_{n=-\infty}^{\infty}z^nq^{n^2}=\prod_{n\geq 0}(1-q^{2n+2})(1+zq^{2n+1})(1+z^{-1}q^{2n+1})$$
Here is the first identity to derive:
$$\sum_{n=-\infty}^{\infty}(-1)^nq^{n^2}=\prod_{m\geq 1}\frac{(1-q^m)}{(1+q^m)}$$
I have gotten to here but am not sure where to go next:
$$\sum_{n=-\infty}^{\infty}(-1)^nq^{n^2}=\prod_{m\geq 1}(1-q^{2m+1})(1-q^{2m})^2\cdot \frac{(1+q^m)}{(1+q^m)}$$
I also need to prove this identity for odd values of $n$:
$$ p(n)=\sum_{m \geq 1}(-1)^{m+1}\left[ p(n-m(2m-1))+p(n-m(2m+1))\right]$$
I know Euler's $p(n)$ formula can be derived from the triple product identity, but how do I make it just the odds?
 A: The first identity: from the Jacobi triple product we have that $$\sum_{n=-\infty}^{\infty}\left(-1\right)^{n}q^{n^{2}}=\prod_{n\geq1}\left(1-q^{2n}\right)\left(1-q^{2n-1}\right)^{2}=
 $$ $$=\prod_{n\geq1}\left(1-q^{2n}\right)\prod_{n\geq1,\, n\, odd}\left(1-q^{n}\right)^{2}=
 $$ $$=\prod_{n\geq1}\left(1-q^{2n}\right)\prod_{n\geq1}\frac{\left(1-q^{n}\right)^{2}}{\left(1-q^{2n}\right)^{2}}=\prod_{n\geq1}\frac{\left(1-q^{n}\right)^{2}}{\left(1-q^{2n}\right)}=
 $$ $$\prod_{n\geq1}\frac{\left(1-q^{n}\right)^{2}}{\left(1+q^{n}\right)\left(1-q^{n}\right)}=\prod_{n\geq1}\frac{\left(1-q^{n}\right)}{\left(1+q^{n}\right)}.$$ I'm working on the second. 
A: For the second problem we can replace $q$ with $q^{2}$ and set $z=-q$ in Jacobi's Triple Product Identity to get $$\dfrac{{\displaystyle \prod_{n=1}^{\infty}(1-q^{n})}}{{\displaystyle \prod_{n=1}^{\infty}\left(1-q^{2(2n-1)}\right)}}= \sum_{n = -\infty}^{\infty}(-1)^{n}q^{n(2n+1)}$$ and note that $$\sum_{n=0}^{\infty}p(n)q^{n}=\left\{\prod_{n=1}^{\infty}(1-q^{n})\right\}^{-1}$$ so that $$\sum_{n=0}^{\infty}p(n)q^{n}\sum_{n=-\infty}^{\infty}(-1)^{n}q^{n(2n+1)}=\prod_{n=1}^{\infty}\left(1-q^{2(2n-1)}\right)^{-1}$$ and therefore the coefficients of odd powers of $q$ are zero in LHS. Finding the expression of coefficient of $q^{n}$ with $n$ odd gives us the desired recursion formula.
