complex analysis - Show $cos pi z=prodlimits_{-infty}^infty left(1-frac{2z}{2n-1}right) e^{frac{2z}{2n-1}}$

I have already shown $\cos \pi z=\prod\limits_{n=1}^{\infty}\left(1-\frac{4z^2}{(2n-1)^2}\right)$. Here is what I have after that,

\begin{equation*}\begin{split}
\cos(\pi z) & = \prod_{n=1}^{\infty}\left(1-\frac{2z}{(2n-1)}\right)\left(1+\frac{2z}{(2n-1)}\right)\\
& = \prod_{n=1}^{\infty}\left(1-\frac{2z}{(2n-1)}\right)\prod_{n=1}^{\infty}\left(1+\frac{2z}{(2n-1)}\right)\\
& = \prod_{n=1}^{\infty}\left(1-\frac{2z}{(2n-1)}\right)\prod_{-\infty}^{n=-1}\left(1-\frac{2z}{(2n+1)}\right)\\
& = \prod_{n=1}^{\infty}\left(1-\frac{2z}{(2n-1)}\right)\prod_{-\infty}^{n=0}\left(1-\frac{2z}{(2n-1)}\right)\\
& = \prod_{n=-\infty}^{\infty}\left(1-\frac{2z}{(2n-1)}\right)
\end{split}\end{equation*}
I'm not sure where the exponential comes from.

Answer

We have $$

\begin{align}
\csc\left(z\right)-\frac{1}{z}= &\sum_{k\in\mathbb{Z},\, k\neq0}\left(-1\right)^{k}\left(\frac{1}{k\pi}+\frac{1}{z-k\pi}\right) \\ = & \sum_{k\in\mathbb{Z},\, k\neq0}\left(\frac{1}{2k\pi}+\frac{1}{z-2k\pi}\right)-\sum_{k\in\mathbb{Z}}\left(\frac{1}{\left(2k-1\right)\pi}+\frac{1}{z-\left(2k-1\right)\pi}\right)
\end{align}$$ so if we integrate we have $$
\begin{align}
\left.\log\left(\frac{\tan\left(z/2\right)}{z}\right)\right|_{0}^{2x}= & \sum_{k=-\infty}^{\infty}\left.\left(\log\left(z-2k\pi\right)+\frac{z}{2k\pi}\right)\right|_{0}^{2x} \\ - & \sum_{k=-\infty}^{\infty}\left.\left(\log\left(z-\left(2k-1\right)\pi\right)+\frac{z}{\left(2k-1\right)\pi}\right)\right|_{0}^{2x}
\end{align}
$$ then $$\frac{\sin\left(x\right)}{x\cos\left(x\right)}=\frac{\prod_{k=-\infty}^{\infty}\left(1-\frac{x}{\pi k}\right)\exp\left(\frac{x}{k\pi}\right)}{\prod_{k=-\infty}^{\infty}\left(1-\frac{2x}{\pi\left(2k-1\right)}\right)\exp\left(\frac{2x}{\pi\left(2k-1\right)}\right)}.
$$ With the same method, from $$\cot\left(z\right)-\frac{1}{z}=\sum_{k\in\mathbb{Z},\, k\neq0}\left(\frac{1}{k\pi}+\frac{1}{z-k\pi}\right)
$$ it is possible to prove that $$\frac{\sin\left(x\right)}{x}=\prod_{k=-\infty}^{\infty}\left(1-\frac{x}{\pi k}\right)\exp\left(\frac{x}{\pi k}\right)
$$ (and we recall that the term $k=0

$ is omitted) hence $$\cos\left(x\right)=\prod_{k=-\infty}^{\infty}\left(1-\frac{2x}{\pi\left(2k-1\right)}\right)\exp\left(\frac{2x}{\pi\left(2k-1\right)}\right).
$$