The Riemann zeta function $\zeta(s)$ has no zeros for $\Re(s)>1$ I write $\zeta(s)$ for $\Re(s)>1$ as:

$\zeta(s) = \prod_{p} (1-p^{-s})^{-1}$

Using this I can show that the Riemann zeta function has no zero for $\Re(s)>1$. I'm however not sure about the next step. I would like to use the zero product property, but I know it doesn't necessary hold for infinite products. How do I handle this situation? 
 A: The following proposition can be found in 'Complex Analysis' by Stein and Shakarchi (Pg. 141): If $\sum_n |a_n|$ converges then the product
$$ \prod_{n=1}^{\infty}(1+a_n) $$ 
converges and in this case the product converges to 0 if and only if one of the factors is 0. In this case we have
\begin{align*}
(1-p^{-s})^{-1}&=\left(\frac{p^s-1}{p^s}\right)^{-1} \\
&=\frac{p^s}{p^s-1} \\
&=1+\frac{1}{p^s-1}
\end{align*}
so apply the proposition for $a_n=(p^s-1)^{-1}$ to see that the product converges for Re$(s)>1$. Then we know $(1-p^{-s})^{-1}\neq 0$ for all primes $p$ and so using the identity and the second statement in the proposition $\zeta(s)\neq 0$ for all $s\in\mathbb{C}$ with Re$(s)>1$.
A: It does hold (for absolutely convergent products, see below): just like terms of a sum necessarily need to tend to 0 for the sum to converge, terms of a product need to tend to 1 for it to converge. So you would find zeros in the "finite" factors of the product. 
Edit: As pointed out in the comments, you need to be a bit careful with infinite products. However, the statement is true if you consider absolutely convergent products, which is all we need for $\zeta(s)$.
