The infinite product of $sin(\pi z)$ is said to be...
$\sin \pi z=\pi z\prod _{{n\neq 0}}\left(1-{\frac {z}{n}}\right)e^{{z/n}}=\pi z\prod _{{n=1}}^{\infty }\left(1-{\frac {z^{2}}{n^{2}}}\right)$
Based on the elementary factors provided by the Weierstrass factorization theorem...
$E_{n}(z)={\begin{cases}(1-z)&{\text{if }}n=0,\\(1-z)\exp \left({\frac {z^{1}}{1}}+{\frac {z^{2}}{2}}+\cdots +{\frac {z^{n}}{n}}\right)&{\text{otherwise}}.\end{cases}}$
I can see why we have...
$\sin \pi z=\pi z\prod _{{n\neq 0}}\left(1-{\frac {z}{n}}\right)e^{{z/n}}$
But I don't understand how the second part was derived. Where did the exponent go? Can someone explain why the following is true?
$\pi z\prod _{{n\neq 0}}\left(1-{\frac {z}{n}}\right)e^{{z/n}}=\pi z\prod _{{n=1}}^{\infty }\left(1-{\frac {z^{2}}{n^{2}}}\right)$
You pair together the $n$ and $-n$ terms on the infinite product.
Note that $(1+\frac{z}{n})(1+\frac{z}{-n})e^{\frac{z}{n}}e^{\frac{z}{-n}} = (1-\frac{z^2}{n^2})$
Thus we can change the index of the product from $n\neq 0$ to $n=1$ to $\infty$ by replacing the $n$th term as above.