TITLE: Trying to get the infinite product for $\sin x$
QUESTION [8 upvotes]: I start with the fact that the zeros of $\sin x$ are $ n\pi$, $n\in\mathbb{Z}$. Therefore, it should be possible to express it as an infinite product: $$\sin x  = x (x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\cdots$$ $$ = x\prod_{n=1}^\infty (x^2 - n^2\pi^2) $$
From here I cannot factor out $\pi^2$ as there are infinite terms and $\pi^\infty$ doesn't look reasonable. If I take out $n^2$ then I have $$x(n!)^2 \prod_{n=1}^\infty (\frac{x^2}{n^2} - \pi^2)$$
But this isn't complete. Infact, this looks wrong (n is defined inside the product afterall). What further manipulations would enable me to get the correct product: $$\sin x  = x\prod(1-\frac{x^2}{n^2\pi^2})$$

REPLY [6 votes]: The infinite product representation for the sine function is
$$\sin(\pi x)=\pi x\prod_{1}^\infty \left(1-\frac{x^2}{n^2}\right).$$
So in the post, $\sin x$ should be replaced by $\sin(\pi x)$. Then the issue raised in the post disappears. 
To prove the result, one needs quite a bit more function theory than the informal type of reasoning about zeros.  Note that for example $e^z$ has no zeros, but $e^z$ is not equal to the empty product. 
Your informal attack on the infinite product is not very distant from that of Euler. That is very good company to be in!