TITLE: A simple way to obtain $\prod_{p\in\mathbb{P}}\frac{1}{1-p^{-s}}=\sum_{n=1}^{\infty}\frac{1}{n^s}$.
QUESTION [31 upvotes]: 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.

REPLY [2 votes]: 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 $\zeta (s)$.