Solving Basel Problem using euler infinite product and infinite product-sum equality Trying to prove Basel problem through the equality $sin(x) = x\prod\limits_{k=1}^{+\infty}(1-\frac{x^{2}}{\pi^{2}k^{2}})$,
I came across the following problem;
I was able to prove the following equality by induction in a finite case, I'd like to prove the general one,which is,for $\{a_{k}\} \subseteq \mathbb{R}$ :
$$\prod_{k \in I}(1+a_{k}) = \sum\limits_{n \in \mathbb{N}}(\sum\limits_{\underset{J \in F(\mathbb{N_{+}})}{\lvert J \rvert = n}} \prod_{k \in J}a_{k})$$
Where $F(\mathbb{N_{+}})$ denotes the finite subsets on $\mathbb{N_{+}}$.
Any tip,suggestion or sketch of the proof would be appreciated.
 A: Note that I just rephrase Proof 1 from here (http://math.cmu.edu/~bwsulliv/basel-problem.pdf).
Let $iy = x/\pi$. Then we have $\sin(i\pi y) = i \pi y\Pi_{k = 1}^{\infty}(1 + y^2/k^2)$. Now let's apply logarithm on both sides to get $\log (\sin(i\pi y)) = \log(i\pi y) + \Sigma_{k = 1}^{\infty}\log(1 + y^2/k^2)$. Taking the derivative with respect to $y$ on both sides yields: $$\frac{i \pi \cos(i \pi y)}{\sin(i\pi y)} = \frac{1}{y} + \Sigma_{k = 1}^{\infty}\frac{2y}{k^2(1 + y^2/k^2)} = \frac{1}{y} + \Sigma_{k = 1}^{\infty}\frac{2y}{k^2 + y^2}$$$$ \implies\Sigma_{k = 1}^{\infty}\frac{1}{k^2 + y^2} = -\frac{1}{2y^2} + \frac{i \pi \cos(i \pi y)}{2y\sin(i\pi y)}$$
So if we can find $lim_{y \rightarrow 0}RHS$ we solve the problem. Note that here I use my own derivation instead of relying to the slides mentioned above. In the following steps we use L'Hôpital's rule twice: 
$$\lim_{y \rightarrow0}-\frac{1}{2y^2} + \frac{i \pi \cos(i \pi y)}{2y\sin(i\pi y)} = \lim_{y \rightarrow0}\frac{i\pi y\cos(i\pi y) - \sin(i\pi y)}{2y^2\sin(i \pi y)} \overset{BH}{=} \lim_{y \rightarrow0}\frac{i\pi \cos(i\pi y) + \pi^2 y\sin(i\pi y) - i \pi \cos(i\pi y)}{4y\sin(i \pi y) + 2i\pi y^2\cos(i\pi y)} = \lim_{y \rightarrow0}\frac{\pi^2 \sin(i\pi y)}{4\sin(i \pi y) + 2i\pi y\cos(i\pi y)} \overset{BH}{=} \lim_{y \rightarrow0}\frac{i\pi^3\cos(i\pi y)}{4i\pi \cos(i \pi y) + 2i\pi \cos(i\pi y) + 2\pi^2 y\sin(i\pi y)} = \frac{i\pi^3}{6i\pi} = \frac{\pi^2}{6}$$.
