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I'd like to prove(explicilty compute) the following relation \begin{align} \prod_{m,n\in \mathbf{Z}}(m+n\tau+u)=iy^{\frac{1}{2}} q^{\frac{1}{12}} \prod_{n=1}^{\infty} (1-yq^n)(1-y^{-1}q^{n-1}) \end{align}

First, i start with the identity that i asked before $\prod_{n\in \mathbf{Z}}(n+a) = e^{i\pi a}-e^{-i\pi a}$

\begin{align} \prod_{m,n\in \mathbf{Z}}(m+n\tau+u) &= \prod_{m\in \mathbf{Z}} (m+u) \prod_{n=1}^{\infty} (m+ n\tau + u)(m-n\tau + u) \\ &= (e^{i\pi u} - e^{-i\pi u}) \prod_{n=1}^{\infty} (e^{i \pi(n\tau +u )} -e^{-i\pi (n\tau+u)})(e^{i \pi(-n\tau +u )} -e^{-i\pi (-n\tau+u)}) \\ &= (e^{i\pi u} - e^{-i\pi u}) \prod_{n=1}^{\infty} e^{-i\pi (n\tau+u)}e^{i\pi (-n\tau +u)}(e^{2i\pi (n\tau+u)}-1)(1-e^{-2i\pi (-n\tau+u)}) \\ & = (y^{\frac{1}{2}} - y^{-\frac{1}{2}}) \prod_{n=1}^{\infty}(-1) q^{-n} (1-yq^n)(1-y^{-1}q^{n}) \end{align}

Where i used $q=e^{2\pi i \tau}$, $y=e^{2\pi i u}$.

I have been calculated up to this(up to far it seems trivial), but the next line is problem to me

\begin{align} =iy^{\frac{1}{2}} q^{\frac{1}{12}} \prod_{n=1}^{\infty} (1-yq^n)(1-y^{-1}q^{n-1}) \end{align}

I have problem with last term $q^{n-1}$ and the first term $y^{\frac{1}{2}}$ I guess the term $q^{\frac{1}{12}}$ comes from zeta function regularization $\prod_{n=1}^{\infty}q^n=q^{-\frac{1}{12}}$ and factor $i$ comes from $ \prod_{n=1}^{\infty} (-1) = (-1)^{\frac{1}{2}} = i^{-1}=-i$

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\begin{align} (y^{\frac{1}{2}} - y^{-\frac{1}{2}}) \prod_{n=1}^{\infty}(-1) q^{-n} (1-yq^n)(1-y^{-1}q^{n}) =iy^{\frac{1}{2}} q^{\frac{1}{12}} \prod_{n=1}^{\infty} (1-yq^n)(1-y^{-1}q^{n-1}) \end{align}

In the process i used \begin{align} (y^{\frac{1}{2}} - y^{-\frac{1}{2}}) \prod_{n=1}^{\infty} (1-y^{-1}q^{n}) = y^{\frac{1}{2}}\prod_{n=1}^{\infty} (1-y^{-1}q^{n-1}) \end{align} Note $n=1$ gives the $(y^{\frac{1}{2}} - y^{-\frac{1}{2}})$ term exactly!.

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