= Solution
One form of the <Jacobi triple product> is, for $0<|q|<1$ and $z\ne0$,
$$
\boxed{\sum_{n\in\mathbb Z}q^{n^2}z^n
=\prod_{j=1}^\infty(1-q^{2j})(1+zq^{2j-1})(1+z^{-1}q^{2j-1}).}
$$
Both sides converge locally uniformly on the punctured $z$ plane. Here is a finite-product proof which also determines the multiplicative constant.
Put $Q=q^2$ and $(Q;Q)_m=\prod_{j=1}^m(1-Q^j)$, with value one at $m=0$. Define the <Gaussian binomial coefficient> ${m\brack k}_Q=(Q;Q)_m/[(Q;Q)_k(Q;Q)_{m-k}]$. The recurrence ${m\brack k}_Q={m-1\brack k}_Q+Q^{m-k}{m-1\brack k-1}_Q$ proves by induction that
$$
\prod_{j=0}^{m-1}(1+tQ^j)
=\sum_{k=0}^mQ^{k(k-1)/2}{m\brack k}_Qt^k.
$$
Rewrite the two finite factors as
$$
\prod_{j=1}^N(1+zq^{2j-1})(1+z^{-1}q^{2j-1})
=z^{-N}q^{N^2}\prod_{j=0}^{2N-1}(1+zq^{1-2N}Q^j).
$$
The finite identity, with $n=k-N$, makes this $\sum_{n=-N}^Nz^nq^{n^2}{2N\brack N+n}_Q$. Multiply by $(Q;Q)_N$. For each fixed integer $n$, the resulting coefficient tends to one, because $(Q;Q)_m$ tends to a nonzero limit and ${2N\brack N+n}_Q\to1/(Q;Q)_\infty$.
All finite $(Q;Q)_m$ are uniformly bounded above and away from zero. Thus these coefficients have one uniform bound, and $\sum_{n\in\mathbb Z}|q|^{n^2}|z|^n$ is uniformly summable on compact annuli. Dominated convergence passes the limit through the finite sums and proves the boxed formula. At $q=0$ the formula follows by its limiting value one. The frequently used alternative form is
$$
\prod_{j\geq1}(1-t^j)(1+u t^{j-1})(1+u^{-1}t^j)
=\sum_{m\in\mathbb Z}u^m t^{m(m-1)/2},\qquad |t|<1.
$$
It follows by the substitution $q^2=t$, $z=u/q$, and is the version used for the charge-counting identity in Question 8.
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