A simple way to obtain $\prod_{p\in\mathbb{P}}\frac{1}{1-p^{-s}}=\sum_{n=1}^{\infty}\frac{1}{n^s}$. 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
$$
\prod_{p\in\mathbb{P}}\frac{1}{1-p^{-s}}=\sum_{n=1}^\infty \frac{1}{n^s}
$$
But I would like to know if there is a simple and elegant way to achieve this result is up through the above list.
 A: 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.
A: 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
$$
P\left(\bigcap_{i=1}^n E_{p_i}\right) = P(E_{\prod_{i=1}^np_i}) = \left(\prod_{i=1}^n p_i \right)^{-s} = \prod_{i=1}^n p_i^{-s} = \prod_{i=1}^n P(E_{p_i})
$$
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
$$
\frac{1}{\zeta(s)} = P(X=1) = P\left(\bigcap_p E_p^c\right) = \prod_p(1-P(E_p)) = \prod_p(1-p^{-s}).
$$
A: I understand that this question is quite old and maybe you are already aware of this proof, but since your profile still says that you would like to have some elegant proofs, here's one simple sieving technique that I absolutely adore and find worth sharing on a platform like MSE-
$$\zeta(s)=1+\frac 1{2^s}+\frac 1{3^s}+\frac 1{4^s}+\frac 1{5^s}+\dots$$
So,
$$\frac 1{2^s}\zeta(s)=\frac 1{2^s}+\frac 1{4^s}+\frac 1{6^s}+\frac 1{8^s}+\dots$$
Subtracting the second equation from the first one, we have
$$\left(1-\frac 1{2^s}\right)\zeta(s)=1+\frac 1{3^s}+\frac 1{5^s}+\frac 1{7^s}+\frac 1{9^s}+\dots$$
So,
$$\frac 1{3^s}\left(1-\frac 1{2^s}\right)\zeta(s)=\frac 1{3^s}+\frac 1{9^s}+\frac 1{15^s}+\frac 1{21^s}+\frac 1{27^s}+\dots$$
Subtracting the fourth equation from the third one, we have
$$\left(1-\frac 1{3^s}\right)\left(1-\frac 1{2^s}\right)\zeta(s)=1+\frac 1{5^s}+\frac 1{7^s}+\frac 1{7^s}+\frac 1{11^s}+\frac 1{13^s}+\frac 1{17^s}+\dots$$
It is clearly visible that this seiving technique can be continued to have
$$\dots \left(1-\frac 1{11^s}\right)\left(1-\frac 1{7^s}\right)\left(1-\frac 1{5^s}\right)\left(1-\frac 1{3^s}\right)\left(1-\frac 1{2^s}\right)\zeta(s)=1$$
which gives
$$\zeta(s)=\frac 1{\left(1-\frac 1{2^s}\right)\left(1-\frac 1{3^s}\right)\left(1-\frac 1{5^s}\right)\left(1-\frac 1{7^s}\right)\left(1-\frac 1{11^s}\right)\dots}$$
In other words,
$$\zeta(s)=\prod_{p\in \mathbb P}\frac 1{1-p^{-s}}$$
which completes the proof.
To make this proof rigorous, we need only to observe that when $\mathfrak R(s)>1$, the sieved RHS approaches $1$, a fact which is a consequence of the convergence of Dirichlet series of $\zeta (s)$.
