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At one point Euler assumes that $$\frac{\sin x}{x} = \prod_{n=1}^{\infty}\left(1-\frac{x}{n\pi} \right)\left(1-\frac{-x}{n\pi} \right)$$ Why does he assume that? If we factor random functions according to their roots, we can conclude $$e^x \frac{\sin x}{x}= \prod_{n=1}^{\infty}\left(1-\frac{x}{n\pi} \right)\left(1-\frac{-x}{n\pi} \right)=\frac{\sin x}x$$

Rasmus Erlemann
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    look into his book introductio in analysin infinitorum (english translation exists); he explains very well why this second formula fails (the terms don't diminish fast enough - he used rather an "infinite number" that he denoted $i$, but his argument can be translated to a convergence argument) – user8268 Jan 23 '15 at 10:40
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    There's more to it than just the zeros. I don't know how Euler did it, but one can look at the logarithmic derivatives, and the value at certain points, to determine the non-vanishing factor in a product representation. – Daniel Fischer Jan 23 '15 at 10:40

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