Here is my idea.
$\sinh \pi z=\frac{1}{2}(e^{\pi z}-e^{-\pi z})$, its zeros are $a_n=ni$, and $a_0$ is a zero of order $1$. Since $\forall R>0,$ $\sum_{n=1}^{\infty}\left(\frac{R}{|a_n|}\right)^2$ converges, so we can write $\sinh \pi z$ as $$\sinh \pi z=ze^{\phi(z)}\prod _{n\in \mathbb{Z}\\n\neq 0}(1-\frac{z}{ni })e^{\frac{z}{ni}}=ze^{\phi(z)}\prod _{n=1}^{\infty}(1+\frac{z^2}{n^2}).$$
We can concldued that $e^{\phi(0)}=\pi$, but how to see $e^{\phi(z)}=\pi$? Can anyone help me? Thank you very much!