The Riemann zeta function $\zeta(s)$ has no zeros for $\Re(s)>1$ [duplicate]

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?

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$(s)>1$.