Extending the Riemann zeta function using Euler's Theorem. Euler's theorem states that if the real part of a complex number $z$ is larger than 1, then $\zeta(z)=\displaystyle\prod_{n=1}^\infty \frac{1}{1-p_n^{-z}}$, where $\zeta(z)=\displaystyle\sum_{n=1}^\infty n^{-z}$ is the Riemann zeta function and $\{p_n\}$ is the set of primes in increasing order.
After reading the argument on extending $\zeta(z)$ to a meromorphic function for those complex numbers $z$ whose real part is larger than $0$, I am wondering whether it is possible to write 
$\zeta(z)=\displaystyle\prod_{n=1}^\infty f_n(z)\cdot\frac{1}{1-p_n^{-z}}, \hspace{0.1in} 0< \Re z <1,$
for some holomorphic functions $f_n$.  I am new to the Riemann zeta function, so if this is a silly question forgive me. Thank you.

Yes and no, if you want something nontrivial. The analytically extended zeta function admits a Weierstrass factorization (called the Hadamard factorization in this case):
$$\zeta(s)=\pi^{s/2}\frac{\prod_{\rho}(1-s/\rho)}{2(s-1)\Gamma(1+s/2)},$$
where the product is over the nontrivial zeros. You can also see the trivial zeros at $s=-2k$ from the Gamma function. You can further expand the product via:
$$\Gamma(1+s/2)=\frac{e^{-\gamma z}}{z}\prod_{k=1}^\infty \left(1+\frac{1+s/2}{k}\right)^{-1}e^{(s+1/2)/k}.$$
This is about as good a product as you'll get and it'll agree with the Euler formula for $s>1$. 
