Proving $~\prod~\frac{\cosh\left(n^2+n+\frac12\right)+i\sinh\left(n+\frac12\right)}{\cosh\left(n^2+n+\frac12\right)-i\sinh\left(n+\frac12\right)}~=~i$ How could we prove that

$${\LARGE\prod_{\Large n\ge0}}~\frac{\cosh\left(n^2+n+\dfrac12\right)+i\sinh\left(n+\dfrac12\right)}{\cosh\left(n^2+n+\dfrac12\right)-i\sinh\left(n+\dfrac12\right)}~=~i$$


This is problem $A12-1$ from the Harvard College Mathematics Review, Volume $4$, Spring $2012$, Chapter $9$, Page $77$, proposed by Moubinool Omarjee, Paris, France.

Needless to say, despite thinking on and off about it for the past couple of weeks, I haven’t yet been able to crack it, partly due to the fact that the argument of $\cosh$ is different than that of $\sinh$. I’ve also noticed that, by completing the square, we have $n^2+n+\dfrac12=\left(n+\dfrac12\right)^2+\left(\dfrac12\right)^2$. That’s about it, I’m afraid.
 A: $$\begin{align}\frac{\cosh\left(n^2+n+\dfrac12\right)+i\sinh\left(n+\dfrac12\right)}{\cosh\left(n^2+n+\dfrac12\right)-i\sinh\left(n+\dfrac12\right)} &= \exp \left(2i\tan^{-1} \frac{\sinh \left(n+\frac{1}{2}\right)}{\cosh \left(n^2+n+\frac{1}{2}\right)}\right)\\&= \exp \left(2i\tan^{-1} \frac{e^{-n^2} - e^{-(n+1)^2}}{1+e^{-n^2}e^{-(n+1)^2}}\right)\\&= \exp{2i\left(\tan^{-1} e^{-n^2} - \tan^{-1} e^{-(n+1)^2}\right)}\end{align}$$
Thus, $$\prod_{n \ge 0}\frac{\cosh\left(n^2+n+\dfrac12\right)+i\sinh\left(n+\dfrac12\right)}{\cosh\left(n^2+n+\dfrac12\right)-i\sinh\left(n+\dfrac12\right)} = \exp{(2i\tan^{-1} 1)} = i$$
via telescopic summation.
A: Another (similar) way is to use this identity $$ \frac{\cosh\left(n^{2}+n+\frac{1}{2}\right)+i\sinh\left(n+\frac{1}{2}\right)}{\cosh\left(n^{2}+n+\frac{1}{2}\right)-i\sinh\left(n+\frac{1}{2}\right)}=\frac{\left(e^{n^{2}}+i\right)\left(e^{\left(n+1\right)^{2}}-i\right)}{\left(e^{n^{2}}-i\right)\left(e^{\left(n+1\right)^{2}}+i\right)} \tag{1}
 $$ to get $$\prod_{n\geq0}\frac{\cosh\left(n^{2}+n+\frac{1}{2}\right)+i\sinh\left(n+\frac{1}{2}\right)}{\cosh\left(n^{2}+n+\frac{1}{2}\right)-i\sinh\left(n+\frac{1}{2}\right)}=\prod_{n\geq0}\frac{\left(e^{n^{2}}+i\right)\left(e^{\left(n+1\right)^{2}}-i\right)}{\left(e^{n^{2}}-i\right)\left(e^{\left(n+1\right)^{2}}+i\right)}=\frac{1+i}{1-i}=i.
 $$ For proving $(1)$ note that $$\frac{\cosh\left(n^{2}+n+\frac{1}{2}\right)+i\sinh\left(n+\frac{1}{2}\right)}{\cosh\left(n^{2}+n+\frac{1}{2}\right)-i\sinh\left(n+\frac{1}{2}\right)}=\frac{\cosh\left(n^{2}+n+\frac{1}{2}\right)2e^{n^{2}+n+\frac{1}{2}}+ie^{n^{2}}\sinh\left(n+\frac{1}{2}\right)2e^{n+\frac{1}{2}}}{\cosh\left(n^{2}+n+\frac{1}{2}\right)2e^{n^{2}+n+\frac{1}{2}}-ie^{n^{2}}\sinh\left(n+\frac{1}{2}\right)2e^{n+\frac{1}{2}}}=
 $$ $$=\frac{e^{2n^{2}+2n+1}+1+ie^{n^{2}}\left(e^{2n+1}-1\right)}{e^{2n^{2}+2n+1}+1-ie^{n^{2}}\left(e^{2n+1}-1\right)}=\frac{e^{2n^{2}+2n+1}-ie^{n^{2}}+ie^{n^{2}+2n+1}+1}{e^{2n^{2}+2n+1}+ie^{n^{2}}-ie^{n^{2}+2n+1}+1}
 $$ $$=\frac{\left(e^{n^{2}}+i\right)\left(e^{\left(n+1\right)^{2}}-i\right)}{\left(e^{n^{2}}-i\right)\left(e^{\left(n+1\right)^{2}}+i\right)}.
 $$
