The evaluation of the infinite product $\prod_{k=2}^{\infty} \frac{k^{2}-1}{k^{2}+1}$

How does one show that$$ \prod_{k=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} =\frac{\pi}{\sinh \pi} ?$$

My attempt: $$ \begin{align} \prod_{k=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} &= \lim_{n \to \infty} \prod_{k=2}^{n}\frac{(k-1)(k+1)}{(k-i)(k+i)} \\ &= \lim_{n \to \infty} \frac{\Gamma(n) \Gamma(n+2) \Gamma(2-i) \Gamma(2+i)} {2 \Gamma(n+1-i) \Gamma(n+1+i)} \\ &= \lim_{n \to \infty} \frac{\Gamma(n)\Gamma(n+2) (1-i) \Gamma(1-i) (1+i)i \Gamma(i)}{2 \Gamma(n+1-i)\Gamma(n+1+i)} \\ &= \frac{\pi}{\sinh \pi}\lim_{n \to \infty} \frac{\Gamma(n)\Gamma(n+2)}{\Gamma(n+1-i) \Gamma(n+1+i)} \end{align}$$

I'm not sure how to go about showing that the limit evaluates to $1$.

EDIT:

To evaluate that limit we can use the fact that $ \displaystyle \frac{\Gamma(n)}{\Gamma(n+z)} \sim n^{-z}$ as $ n \to \infty$.

$$ \begin{align} \lim_{n \to \infty} \frac{\Gamma(n)\Gamma(n+2)}{\Gamma(n+1-i) \Gamma(n+1+i)} &= \lim_{n \to \infty} \frac{\Gamma(n) (n+1)n \Gamma(n)}{(n-i)\Gamma(n-i) (n+i)\Gamma(n+i)} \\ &= \lim_{n \to \infty} \frac{\Gamma(n)n^{-i}}{\Gamma(n-i)} \frac{\Gamma(n) n^{i}}{\Gamma(n+i)} \frac{n^2+n}{n^{2}+1} \\ &= (1)(1)(1) \\ &= 1 \end{align}$$

If you know the product representation

$$\sin (\pi z) = \pi z \prod_{k = 1}^\infty \left(1 - \frac{z^2}{k^2}\right),\tag{1}$$

it is rather easy.

Setting $z = i$, we obtain

$$\frac{\sin (\pi i)}{\pi i} = \frac{\sinh \pi}{\pi} = \prod_{k=1}^\infty \left(1 - \frac{i^2}{k^2}\right) = \prod_{k=1}^\infty \left(1 + \frac{1}{k^2}\right) = 2 \prod_{k=2}^\infty \left(\frac{k^2+1}{k^2}\right).$$

On the other hand,

$$\prod_{k=2}^n \left(\frac{k^2-1}{k^2}\right) = \frac12\cdot \frac32\cdot \frac23\cdot \frac43 \dotsb \frac{n-1}{n}\cdot \frac{n+1}{n} = \frac12\cdot\frac{n+1}{n},$$

so

$$\prod_{k=2}^\infty \left(\frac{k^2-1}{k^2}\right) = \frac12.$$

Now divide.