In Euler's proof he uses the formula:
$$\sin z = z \prod_{n \mathop = 1}^\infty \left({1 - \frac {z^2} {n^2 \pi^2}}\right)$$
and compares coefficients of the $z^3$ term in the Maclaurin series of Sine.
But I was wondering how I can justify (rigorously) that the infinite product can be expanded like
$$x - \left( \frac 1 {\pi^2} + \frac 1 {4 \pi^2} + \cdots \right)x^3 + ...$$
I understand how it works using Newton's identities but I'm not sure how to deal with the fact that it is an infinite product.
Suppose that you consider $$P_k= \prod_{n \mathop = 1}^{k} \left({1 - \frac {z^2} {n^2 \pi^2}}\right)$$ Now, develop as a Taylor series limited to second order $\big(P_k-P_{k-1}\big)$and get $$P_k-P_{k-1}=-\frac{z^2}{k^2 \pi ^2}+O\left(z^4\right)$$ So, the coefficient of $z^2$ is given by $$-\frac{1}{\pi^2}\sum_{i=1}^{i=k}\frac{1}{i^2}=-\frac{H_k^{(2)}}{\pi^2}$$ and the limit, for an infinite value of $k$, of the above harmonic number is equal to $\frac{\pi^2}{6}$.