How to prove the identity $\pi^{s/2}=e^{(\log(2\pi)-1-\gamma/2)s}\prod_{\rho}e^{s/\rho} $? In the wikipedia the Hadamard product for the Riemann's zeta function has two forms. The first one is
$$\zeta(s)=\frac{e^{(\log(2\pi)-1-\gamma/2)s}}{2(s-1)\Gamma(1+s/2)}\prod_{\rho}\left(1-\frac{s}{\rho} \right)e^{s/\rho}$$
The second one is simplified to the form
$$\zeta(s)=\frac{\pi^{s/2}}{2(s-1)\Gamma(1+s/2)}\prod_{\rho}\left(1-\frac{s}{\rho} \right)$$
So my question is, how to prove the identity
$$e^{(\log(2\pi)-1-\gamma/2)s}\prod_{\rho}e^{s/\rho}= \pi^{s/2}$$
without even know the nontrivial roots, $\rho$, of $\zeta(s)$?

Edit 1:
This is equivalent to calculate the value of
$$\sum_{\rho}\frac{1}{\rho}$$
But, how?

Use the fact that $Z(n)=\sum_k \rho_k^{-n}$ can be calculated explicitely for $n=1$ to:
$$
Z(1)=\frac12[2+\gamma-\ln(4\pi)]=1+\gamma/2-\log(2\pi^{1/2}),\tag{$\star$}
$$
eq. (5) from here.
Take the log of your equation in question and see that $s$ cancels out:
$$
{(\log(2\pi)-1-\gamma/2)}+\sum_{\rho}{\rho^{-1}}= \log \pi^{1/2}\\
{(\log(2\pi)-1-\gamma/2)}+(1+\gamma/2-\log(2\pi^{1/2}))=\log{\frac{2\pi}{2\pi^{1/2}}}= \frac12 \log \pi.
$$
Unfortunately I don't have a proof for $(\star)$ yet...
