# Euler's Basel problem proof explanation

First question: After using the function's Maclaurin series, we factor it $$\frac{\sin x}x=\sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n}}{(2n+1)!}=\prod_{n=1}^{\infty}\left(1-\frac{x}{n\pi} \right)\left(1-\frac{-x}{n\pi} \right).$$ Why can we do this? What result did Euler use here?

Second question: After sorting out the first question, we get $$\frac{\sin x}{x}= \prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right).$$ Next, he concludes that $$\frac{\sin x}{x}=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\ldots = 1-\left(\frac1{\pi^2}+\frac1{4\pi^2}+\frac{1}{9\pi^2}+\ldots \right)x^2+\ldots.$$ When exactly can we factor infinite product that way? What result is he using here?

I see no reason to insert the Macluarin series. Euler's brilliant idea was to say something like, "hey, if I know all of the complex zeroes of a polynomial, then I can completely factor it. I know the zeroes of $\sin x$, so maybe it makes sense to factor that. There are infinitely many zeroes, so I would be expressing $\sin x$ as an infinite product, but if it makes sense to consider infinite series, then why not products?"
A reasonable first attempt would be to try $\sin x = A x (x- \pi)(x+\pi) (x-2 \pi)(x + 2\pi) \cdots$, but with that, you run into convergence problems (for fixed $x$, the factors grow in magnitude, so unless one of the factors is $0$ their partial products cannot converge). But notice that in the first product that you wrote:
$$\prod_{n=1}^{\infty} \left( 1 - \frac{x}{n \pi} \right) \left( 1 + \frac{x}{n \pi} \right),$$
the zeroes of the factors are exactly the nonzero integer multiples of $\pi$. For a fixed $x$, the factors will be getting closer and closer to $1$, so there is a shot that the partial products converge.
Euler didn't initially get into proving that these products do converge and that infinite product expansions make sense. There was criticism of his initial work to this effect -- it wasn't rigorous. An even bigger criticism is that with a polynomial, we must hunt for complex zeroes to factor it completely. There was the possibility that $\sin x$ would have additional complex zeroes, whatever that might mean.