# simplifying a product formula (similar to Euler's sine product)

Can anyone help me out trying to simplify the left hand side of the below equation to obtain the right hand side?

$$\displaystyle\prod_{\substack{n=-\infty \\n\neq 0}}^\infty\left(1-\frac{z}{n}\right)e^{z/n} = \prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right).$$

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Isn't this as simple as combining the terms on the left side with for $n$ and $-n$? – Thomas Andrews May 27 '12 at 15:22

## 1 Answer

Note that $$\left(1-\frac z n\right) e^{z/n} \cdot \left(1+\frac z n\right) e^{-z/n} = 1-\frac{z^2}{n^2}.$$

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