For $z\in\mathbb{C}$, with $\Re(z)>1:$$$ \zeta(z)=\prod_p \frac1{1-p^{-z}}.$$
Proof:
Let $z\in \mathbb{C}$, with $\Re(z)>1$. Write the prime numbers down as $2=p_1<p_2<p_3<\dots$. Then the given product can be expressed more explicitly as $$ \prod_{n=1}^{\infty}\frac1{1-p_n^{-z}}.$$
We then have
$$ \prod_{k=1}^n\left(1+\frac1{p_k^z}+\frac1{p_k^{2z}}+\frac1{p_k^{3z}}+\dots\right) = \sum_{m_1,\dots,m_n=0}^{\infty}\frac1{(p_1^{m_1}\cdots p_n^{m_n})^z}.$$ The unique prime factorization of natural numbers implies that $$\left| \sum_{k=1}^{\infty}\frac1{k^z}-\sum_{m_1,\dots,m_n=0}^{\infty}\frac1{(p_1^{m_1}\cdots p_n^{m_n})^z}\right|\le \sum_{k=p_n}^{\infty}\frac1{|k^z|}\to 0, \text{as } n\to \infty. $$ This proves that $\zeta(z)$ approaches $\prod_{k=1}^n\left(1+\frac1{p_k^z}+\frac1{p_k^{2z}}+\frac1{p_k^{3z}}+\dots\right)$ for $n\to\infty$. The factors of this product are geometric series, so the result follows.
Questions:
Why do we explicity mention that we order the prime numbers the way we do. Does the convergence depend on this order?
"The unique prime factorization of natural numbers implies that $$\left| \sum_{k=1}^{\infty}\frac1{k^z}-\sum_{m_1,\dots,m_n=0}^{\infty}\frac1{(p_1^{m_1}\cdots p_n^{m_n})^z}\right|\le \sum_{k=p_n}^{\infty}\frac1{|k^z|}\to 0, \text{as } n\to \infty." $$ I don't see where the upper bound comes from, and how does it approach $0$ as $n\to\infty$ (since that would just imply that $p_n$ would be a very large prime number)?
Thanks.