How to compute this infinite product

Let's use the following identity :\begin{aligned}\operatorname{e}^{\operatorname{i}\frac{q\pi}{t}}-\operatorname{e}^{\operatorname{i}\frac{n\pi}{t}}&=2\operatorname{i}\operatorname{e}^{\operatorname{i}\frac{\left(q+n\right)\pi}{2t}}\sin{\left(\frac{q-n}{2t}\pi\right)}\end{aligned}

We have :

\begin{aligned}\prod_{q=1\\ q\neq n}^{t}{\sin{\left(\frac{q-n}{2t}\pi\right)}}&=\prod_{q=1}^{n-1}{\sin{\left(\frac{q-n}{2t}\pi\right)}}\prod_{q=n+1}^{t+n-1}{\sin{\left(\frac{q-n}{2t}\pi\right)}}\prod_{q=t+1}^{t+n-1}{\csc{\left(\frac{q-n}{2t}\pi\right)}}\\ &=\prod_{k=1}^{n-1}{\sin{\left(\frac{\left(n-k\right)-n}{2t}\pi\right)}}\prod_{k=1}^{t-1}{\sin{\left(\frac{\left(n+k\right)-n}{2t}\pi\right)}}\prod_{k=1}^{n-1}{\csc{\left(\frac{\left(t+n-k\right)-n}{2t}\pi\right)}}\\ &=\prod_{k=1}^{n-1}{\left(-\sin{\left(\frac{k\pi}{2t}\right)}\right)}\prod_{k=1}^{t-1}{\sin{\left(\frac{k\pi}{2t}\right)}}\prod_{k=1}^{n-1}{\sec{\left(\frac{k\pi}{2t}\right)}}\\ \prod_{q=1\\ q\neq n}^{t}{\sin{\left(\frac{q-n}{2t}\pi\right)}}&=\left(-1\right)^{n-1}\frac{\sqrt{t}}{2^{t-1}}\prod_{k=1}^{n-1}{\tan{\left(\frac{k\pi}{2t}\right)}}\end{aligned}

And : $$\Large \prod_{q=1\\ q\neq n}^{t}{\operatorname{e}^{\operatorname{i}\frac{\left(q+n\right)\pi}{2t}}}=\operatorname{e}^{-\operatorname{i}\frac{n\pi}{t}}\operatorname{e}^{\operatorname{i}\sum\limits_{q=1}^{t}{\frac{\left(q+n\right)\pi}{2t}}}=\operatorname{e}^{-\operatorname{i}\frac{n\pi}{t}}\operatorname{e}^{\operatorname{i}\frac{2n+t+1}{4}\pi} $$

Thus : $$ \Large\prod_{q=1\\ q\neq n}^{t}{\frac{1}{\operatorname{e}^{\operatorname{i}\frac{q\pi}{t}}-\operatorname{e}^{\operatorname{i}\frac{n\pi}{t}}}}=\left(-t\right)^{n-1}\frac{2^{t-1}}{\sqrt{t}}\operatorname{e}^{\operatorname{i}\frac{n\pi}{t}}\operatorname{e}^{-\operatorname{i}\frac{2n+t+1}{4}\pi}\prod_{k=1}^{n-1}{\frac{\cot{\left(\frac{k\pi}{2t}\right)}}{t}} $$

I may have made a mistake, but the expression above doesn't appear to have a limit.