Paradox of $-i$ seemingly equal to $1$ via the Wallis product for $\pi$ and the Euler sine product Assuming $x$ is a real variable throughout,$$\frac{\sinh(ix)}{i}=\sin(x)$$
$$\frac{\sinh(\pi ix)}{\pi ix} = \frac{\sin(\pi x)}{\pi x}$$
$$\frac{e^{\pi ix}-e^{-\pi ix}}{2\pi ix}=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right)$$
Letting $x=\frac{1}{2}$ yields:
$$\frac{2}{\pi i}=\prod_{n=1}^{\infty}\left(1-\frac{1}{4n^2}\right)$$
$$\frac{1}{i}=-i=\frac{\pi}{2}\prod_{n=1}^{\infty}\left(\frac{4n^2-1}{4n^2}\right)=\frac{\pi}{2}\prod_{n=1}^{\infty}\left(\frac{(2n-1)(2n+1)}{4n^2}\right)$$
The famous Wallis product for $\pi$ tells us:
$$\frac{\pi}{2}=\frac{2\cdot 2}{1\cdot 3}\cdot\frac{4\cdot 4}{3\cdot 5}\cdot\frac{6\cdot 6}{5\cdot 7}\dots=\prod_{n=1}^{\infty}\left(\frac{4n^2}{(2n-1)(2n+1)}\right)$$
Therefore, in a multiplication which I hope to be legitimate, we seem to get:
$$-i=\prod_{n=1}^{\infty}\left(\frac{4n^2}{(2n-1)(2n+1)}\right)\cdot\prod_{n=1}^{\infty}\left(\frac{(2n-1)(2n+1)}{4n^2}\right)=\prod_{n=1}^{\infty}\left(\frac{(2n-1)(2n+1)}{(2n-1)(2n+1)}\right)=1$$
Clearly, $1\neq-i$, so what have I done wrong here? I randomly came  up with this while half asleep, so there may be some trivial error here - perhaps the products cannot be combined like this - but I'm at a loss!
Thanks for any help.

You appear to be making a mistake in the left hand side of your fourth equation. $e^{\frac{1}{2}i \pi}=i$, and $e^{-\frac{1}{2}i \pi}=-i$. Therefore
$$\frac{e^{\frac{1}{2}i\pi}-e^{-\frac{1}{2}i\pi}}{2\pi i\frac{1}{2}}=\frac{2i}{2\pi i}=\frac{2}{\pi}$$
Carrying this correction through gives $1=1$ at the end.
