I found that formula on a paper that Riemann wrote call in English "On the Number of Prime Numbers less than a Given Quantity". In that paper Riemann first introduced his Zeta function. It also says that Euler show that. I will thank somebody that can help me understand the proof of the formula.
$$\prod\frac{1}{1-\frac{1}{p^s}} = \sum \frac{1}{n^s} $$
The first chance I got to know this identity was from Hardy's An Introduction to the Theory of Numbers about fifteen years ago when I was beginning to learn calculus.
As mixedmath commented, the book also says it's the unique prime factorization. But I had no idea how this is actually the UPF until I recognized the basic identity
\[ \frac{1}{1 - p^{-s}} = 1 + p^{-s} + p^{-2s} + \cdots. \]
So the left product looks like \[ ( 1 + 2^{-s} + 2^{-2s} + \cdots) ( 1 + 3^{-s} + 3^{-2s} + \cdots) ( 1 + 5^{-s} + 5^{-2s} + \cdots) \cdots \]
How do you expand these product? You pick one term from each parenthesis (you might want to pick infinitely many 1's), and a result looks like \[ (p_{1}^{k_{1}} p_{2}^{k_{2}} \cdots p_{n}^{k_{n}} )^{-s} = n^{-s} \] for some $n = p_{1}^{k_{1}} p_{2}^{k_{2}} \cdots p_{n}^{k_{n}}$.
Now, given some natural number $n$, what if there are two ways to write it in this product form?
Oh yes. It's the UPF that's Hardy (or Wright) says.
BECAUSE it's unique (a clever guy proved it many years ago), there is a one to one correspondence between $n^{-s}$ and a $(product)^{-s}$ arizing from the infinite product.