How do I prove that $\frac{1}{\sin^2 \pi z}=\frac{1}{\pi^2}\sum_{n=-\infty}^{\infty} \frac{1}{(z-n)^2}$ using infinite products?
I know how to prove this without using infinite products. One way is to show that the function which is a difference between the right-hand-side and left-hand-side has removable singularities at all integers then show that it is bounded.
However, I found a book that this problem is in the chapter about infinite products, but I don't get how to do this. Thank you in advance.
Hint: Consider $$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) \tag{1}$$ and apply $\frac{d^2}{dz^2}\log(\cdot)$ to both sides.