Weierstrass product form How to show the Weierstrass product form of the entire function $f(z)= \sinh z$
This question seem so interesting. I would like to write my some ideas, but I dont want to direct incorrectly. Please help me to learn correctly and explicitly. I asked to question in fact in order to learn the topic precisely. 
 A: Hint: consider the functional relation $$\sinh(z)=-i \sin(iz). $$
EDIT: Here is a more detailed explanation:
I will derive the Weierstrass product for $\sin(z)$, and using the identity above we will get the product for $\sinh (z)$ as well.
The zeros of the entire function $\sin(\pi z)$ are the integers $\mathbb Z$, and the smallest integer $h$ which makes the series $$ \sum_{n \in \mathbb Z \setminus \{0 \}} \frac{1}{n^{h+1}}$$ converge is $h=1$. This number is known as the genus of the entire function $\sin(\pi z)$, and it is needed for applying the Weierstrass factorization theorem effectively.
It follows from the Weierstrass factorization theorem that $$\sin(\pi z)=z e^{g(z)} \prod_{n \in \mathbb Z^*} \left(1-\frac{z}{n} \right) e^{z/n} $$
where $\mathbb Z^*=\mathbb Z \setminus\{0 \}$ and $g(z)$ is an entire function to be determined. Taking the logarithmic derivative of both sides (except at the integers) we find $$ \pi \cot (\pi z)=\frac{1}{z}+g'(z)+ \sum_{n \in \mathbb Z^*} \frac{1}{z-n}+\frac{1}{n} .$$ Consider now the development of $\pi \cot(\pi z)$ into partial fractions (this is another exercise by itself):
$$\pi \cot(\pi z) = \frac{1}{z}+\sum_{n \in \mathbb Z^*} \frac{1}{z-n}+\frac{1}{n}$$ from this it follows that $g'(z) \equiv 0$, thus $g(z)$ is a constant.
Taking the limit of $\frac{\sin(\pi z)}{z}$ as $z \to 0$ we find that $e^{g(z)} \equiv \pi$, thus we have found $$\sin(\pi z)= \pi z \prod_{n \in \mathbb Z^*} \left( 1-\frac{z}{n} \right) e^{z/n} $$
combining the factors $\left(1 - \frac{z}{ \pm n} \right) e^{ \frac{z}{ \pm n}}$ gives the equivalent form $$\sin(\pi z)= \pi z \prod_{n=1}^\infty \left(1-\frac{z^2}{n^2}\right). $$
Now replacing $z \mapsto \frac{z}{\pi}$ gives $$\sin(z)=z \prod_{n=1}^\infty \left(1-\frac{z^2}{n^2 \pi^2} \right) .$$
Thus $$\sinh(z)=-i \sin(iz)= -i \cdot i z \prod_{n=1}^\infty \left(1+\frac{z^2}{\pi^2 n^2} \right)=z \prod_{n=1}^\infty \left(1+\frac{z^2}{\pi^2 n^2} \right) .$$
If you are interested in any more details, you may consult Ahlfors' complex analysis text.
I hope everything is clear now, if not I can try to elaborate further.
