how to show that $e^{\pi z}-1 = \pi z e^{\frac{\pi z}{2}}\prod_1^{\infty}(1+\frac{z^2}{4n^2})$? how to show that $e^{\pi z}-1 = \pi z e^{\frac{\pi z}{2}}\prod_1^{\infty}(1+\frac{z^2}{4n^2})$?
I would want to show that $F := \frac{e^{\pi z}-1}{\pi z e^{\frac{\pi z}{2}}}= \prod_1^{\infty}(1+\frac{z^2}{4n^2}) : = G$. Then I want to show that $\frac{F'}{F}= \frac{G'}{G}$, but I do not have any information on $\frac{G'}{G}$
 A: HINT.- It is known  the development in infinite product of hyperbolic sine
$$\sinh z=z\prod_1^{\infty}\left(1+\frac{z^2}{n^2\pi^2}\right)$$ 
(if you want to prove this, try to do it).
so we have
$$\sinh \frac{\pi z}{2}=\frac{\pi z}{2}\large\prod_1^{\infty}\left(1+\frac{z^2}{4n^2}\right)\Rightarrow 2\sinh \frac{\pi z}{2}\large e^{\frac{\pi z}{2}}=\pi z\space\large e^{\frac{\pi z}{2}}\prod_1^{\infty}\left(1+\frac{z^2}{n^2\pi^2}\right)$$
Because of  $$\sinh \frac{\pi z}{2}=\frac{\large e^{\frac{\pi z}{2}}-e^{-\frac{\pi z}{2}}}{2}$$ you can finish.
Additional hint and a bit of history.- Euler had the ingenious idea of considering the function $\frac{\sin x}{ x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}\cdots$ as an "infinite polynomial" that is not null when $x = 0$ (in fact it is equal to $1$) but has the same roots as the functions sine (all the $n\pi$ for $n\ne 0$) so he got "the equality"
$$\frac{\sin x}{ x}=(1-\frac{x}{\pi})(1+\frac{x}{\pi})(1-\frac{x}{2\pi})(1+\frac{x}{2\pi})\cdots=\prod_1^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right)$$
Bernoulli much criticized this lack of rigor. The fact is either the formula was correct (which is known and many books prove it! ) and extends to complex variable $z$  .
Use now the known formula $$\sinh z=-i\sin(iz)$$ to go to the hyperbolic.
A: Hint and alternative approach: 1st approach is to define a complex logarithm and you can put log on both sides , the product inside the logarithm becomes a series and then you could proceed. 2nd approach. Your fraction on the left side is clearly $2\sinh(\pi z/2) $ . Your product on the right side is the product representation of the same function. Then $F^{'}/ F =G^{'}/G $. And there it is. Be careful though. You need to prove this formula and that the product converges in all compact sets.
