Csdrhrt

Let $ p_1<p_2 <cdots <p_k < cdots $ the increasing list in set $mathbb{P}$ of all prime numbers .
By sum of infinite geometric series we have $sum_{k=0}^infty r^k = frac{1}{1-r}$, for $0<r<1$. For all $s>1$ and $r=frac{1}{p_k^{s}}$ we have
$$
begin{array}{cccccc}
dfrac{1}{1-p_{1}^{-s}}
&
=
&
1+dfrac{1}{(p_1^s)^1}+dfrac{1}{(p_1^s)^2}+dfrac{1}{(p_1^s)^3}+
&
!!cdots!!
&
+dfrac{1}{(p_1^{s})^{alpha_1}}+
&
cdots
\
dfrac{1}{1-p_{2}^{-s}}
&
=
&
1+dfrac{1}{(p_2^s)^1}+dfrac{1}{(p_2^s)^2}+dfrac{1}{(p_2^s)^3}+
&
!!cdots!!
&
+dfrac{1}{(p_2^s)^{alpha_2}}+
&
cdots
\
dfrac{1}{1-p_{3}^{-s}}
&
=
&
1+dfrac{1}{(p_3^s)^1}+dfrac{1}{(p_3^s)^2}+dfrac{1}{(p_3^s)^3}+
&
!!cdots!!
&
+dfrac{1}{(p_3^s)^{alpha_3}}+
&
cdots
\
vdots
&
vdots
&
vdots
&
vdots
&
vdots
&vdots
\
\
vdots
&
vdots
&
vdots
&
vdots
&
vdots
&
vdots
\
dfrac{1}{1-p_{k}^{-s}}
&
=
&
1+dfrac{1}{(p_k^s)^1}+dfrac{1}{(p_k^s)^2}+dfrac{1}{(p_k^s)^3}+
&
!!cdots!!
&
+dfrac{1}{(p_k^s)^{alpha_k}}+
&
cdots
\
vdots
&
vdots
&
vdots
&
vdots
&
vdots
&
vdots
\
end{array}
$$

And the Fundamental Theorem of Arithmetic tells us that every integer $ n> 1$ can be decomposed uniquely as a product
$$
n= p_{i_1}^{alpha_{i_1}}p_{i_2}^{alpha_{i_2}}cdots p_{i_k}^{alpha_{i_k}}
$$
of powers of prime numbers $p_{i_1}< p_{i_2}< cdots < p_{i_k}$ for integers $alpha_{i_1},alpha_{i_2},ldots,alpha_{i_k}geq 1$. Since $
n^s= (p_{i_1}^s)^{alpha_{i_1}}(p_{i_2}^{s})^{alpha_{i_2}}cdots (p_{i_k}^s)^{alpha_{i_k}}$ and
using brute force with I can prove that
$$
prod_{pinmathbb{P}}frac{1}{1-p^{-s}}=sum_{n=1}^infty frac{1}{n^s}
$$
But I would like to know if there is a simple and elegant way to achieve this result is up through the above list.

I absolutely love this result, I literally cannot stop from getting goosebumps and smiling whenever I think about it. It is a proof from probability theory! I learned it in David William's Probability with Martingales, of which it is part of exercise E4.2.

Fix $s>1$ and recall that $zeta(s) = sum_{n in mathbb{N}} n^{-s}$, so we aim to show that $1/zeta(s) = prod_p(1-p^{-s})$ where of course $p$ ranges over the primes.

First, define a probability measure $P$ and an $mathbb{N}$-valued random variable $X$ such that $P(X=n) = n^{-s}/zeta(s)$ (for example take $P({n}) = n^{-s}/zeta(s)$ and $X(omega)=omega$). Let $E_k := {X text{ is divisible by } k}$. We claim that the events $(E_p : p text{ prime})$ are independent. We note that
$$
P(E_k) = sum_{i=1}^infty P(X=ik) = sum_{i=1}^infty frac{(ik)^{-s}}{zeta(s)} = k^{-s} frac{zeta(s)}{zeta(s)} = k^{-s}.
$$

Then if $p_1,ldots,p_n$ are distinct primes we have $$bigcap_{i=1}^n E_{p_i} = E_{prod_{i=1}^np_i},$$ so that
$$
Pleft(bigcap_{i=1}^n E_{p_i}right) = P(E_{prod_{i=1}^np_i}) = left(prod_{i=1}^n p_i right)^{-s} = prod_{i=1}^n p_i^{-s} = prod_{i=1}^n P(E_{p_i})
$$
so our independence claim is proved.
Then we note that $1$ is the unique positive integer which is not a multiple of any prime. Hence
$$
frac{1}{zeta(s)} = P(X=1) = Pleft(bigcap_p E_p^cright) = prod_p(1-P(E_p)) = prod_p(1-p^{-s}).
$$

Let $s$ for which $Re(s)>1$ then, for all $p in mathbb{P}$ we have $$sum_{k=1}^{infty} frac{1}{p^{ks}}=left(1-frac{1}{p^s}right)^{-1}$$

Now let $A(N)$ be the set of all strictly positive numbers such as all prime divisors are at maximum $N$.

Then $$prod_{p in mathbb{P},text{ }p<N}left(1-frac{1}{p^s}right)^{-1}=sum_{n in A(N)}frac{1}{n^s}$$

Obviously ${1,...,N}subset A(N)$, then

$$left|zeta(s)-prod_{p in mathbb{P},text{ }p<N}left(1-frac{1}{p^s}right)^{-1}right|le sum_{n>N}frac{1}{n^{Re{(s)}}}$$

When $N$ goes to infinity it gives the result.