Euler Product formula for Riemann zeta function proof In class we introduced Reimann Zeta function
$$
\zeta (x)=\sum_{n=1}^{+\infty} \frac{1}{n^x} 
$$
And we proved its domain was $D=(1,+\infty)$
Now Euler proved that 
$$
\zeta(x)=\prod_{p\text{ prime}}\frac{1}{1-p^{-x}}
$$
By saying
$$
\zeta(x)=1+\frac{1}{2^x}+\frac{1}{3^x}+...   \\
\zeta(x)(\frac{1}{2^x})=\frac{1}{2^x}+\frac{1}{4^x}+... \\
\zeta(x)(1-\frac{1}{2^x})=1+\frac{1}{3^x}+\frac{1}{5^x}+...
$$
And so on for every prime number.
However this proof isn't a 'rigorous proof' as my professor says. Why is that and how would one prove this rigorously? Any reference would be helpful. 
I have seen on wikipedia that to make the proof rigorous we need to observe $\mathfrak{R}(x)>1$ Is that the real part of x or something else? 
 A: (this is how I'd do it)
consider the formal product and series, then by induction on the $k$th prime : $$\prod_p (1+p^{-x}+p^{-2x}+\ldots) = \sum_n a_n n^{-x}$$
now consider the coefficient $a_1$ : it is clearly $1$, the coefficient $a_2$ : it is clearly $1$, etc. (by the fundamental theorem of arithmetic). 
now do the same with $$F_K(x) = \prod_{p \le K} (1+p^{-x}+p^{-2x}+\ldots) = \sum_{n=1}^\infty a_n(K) n^{-x}$$
then if $n = \prod_i p_i^{e_i}$ : $a_n(K) = 1$ if all the $p_i \le K$, otherwise $a_n(K) = 0$.
clearly $F_K(x)$ is well-defined for any $x > 1$ (it is a finite product), and $\lim_{K \to \infty} F_K(x)$ exists too because the logarithm of the infinite product is $-\sum_p \ln(1-p^{-x})$ which is absolutely convergent since $\ln(1-p^{-x}) \sim -p^{-x}$ and that $\sum_p p^{-x} < \sum_{n=1}^\infty n^{-x}$ which is (absolutely) convergent.
finally, $\zeta(x)- F_K(x) = \sum_{n=1}^\infty |a_n(K)-1| n^{-x} > 0$, it is absolutely convergent, it is decreasing in $K$, and it clearly  $\to 0$ when $K \to \infty$ since every term $\to 0$.
i.e. :
$$\lim_{K \to \infty} F_K(x) = \prod_p \frac{1}{1-p^{-x}} = \zeta(x) \qquad\qquad (\forall \ x > 1)$$
the proof for every $Re(x) > 1$ is a little more complicated, since we don't have monotone convergence of $\zeta(x)-F_K(x)$ to $0$ but only absolute convergence.
A: Eulers formula for the Zeta function is,
$$ \prod_{p \in \mathbb{P}}^{p \le A} \frac1{1- p^{-s}} = \prod_{p \in \mathbb{P}} (\sum_{k=0}^{\infty} p^{-ks}) $$
Infinite sums and products may depend on order. Finite do not. Consider the finite products,
$$ \prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} = \prod_{p \in \mathbb{P}}^{p \le A} \sum_{k=0}^{K} p^{-ks} $$
Product over a sum is the sum over the cartesian products of the products.
$$ \prod_{a \in A} \sum_{b \in B_a} t_{a,b} = \sum_{c \in \prod_{a \in A} B_a}  \prod_{a \in A}t_{a,c_a} $$
where the product of sets $\prod_{a \in A} B_a$ is taken to be a cartesian product.
$$ \prod_{p \in \mathbb{P}}^{p \le A} \sum_{k=0}^{K} p^{-ks} = \sum_{v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\}}  \prod_{p \in P}^{p \le A}p^{-v_ps} = \sum_{v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\}}  (\prod_{p \in P}^{p \le A}p^{-v_p})^{-s}$$
Change the summing variable using,
$$ \sum_{v \in V }  g(f(v)) = \sum_{w \in \{f(v) : v \in V \} }  g(w)$$
which is valid only if f is one to one. This is true by Fundamental theorem of arithmetic, as every number has a unique factorization. This gives,
$$ \prod_{k=0}^{K} \sum_{p \in \mathbb{P}}^{p \le A} p^{-ks}  =  \sum_{n \in \{\prod_{p \in P}^{p \le A}p^{-v_p} : v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\} \} } n^{-s} $$
Define Q by,
$$ Q(A, K) = \{\prod_{p \in P}^{p \le A}p^{-v_p} : v \in \prod_{p \in \mathbb{P}}^{p \le A} \{ 0 .. K\} \}$$
No natural number may have a factor greater than itself. Also the highest power it can have for a prime factor is $N = 2^K$, giving $K = \log_2(N)$. Every number from 1..N must have a unique factorization, and that factorization must be constructed by Q.
$$ \{1 .. N\} \subset Q(N, log_2(N)) $$
The positive natural numbers are given by,
$$ \lim_{A \to \infty, K \to \infty}Q(A,K) = \mathbb{N+} = \{\prod_{p \in P}p^{-v_p} : v \in \prod_{p \in \mathbb{P}} \{ 0 .. \infty\} \}$$ 
Then,
$$ \prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} = \prod_{p \in \mathbb{P}}^{p \le A} (\sum_{k=0}^{K} p^{-ks}) =  \sum_{n \in Q(A, K)}  n^{-s} $$

Taking limits,
$$ \lim_{A \to \infty, K \to \infty} \prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} = \prod_{p \in \mathbb{P}} \frac1{1- p^{-s}} $$
$$ \lim_{A \to \infty, K \to \infty} \sum_{n \in Q(A, K)}  n^{-s} = \sum_{n \in \mathbb{N^+}}  n^{-s} $$
The order of summation is not prescribed by the sum. Infinite sums can have different values depending on order. However,
$$ \sum_{1..n}^{\infty}  n^{-s} $$
is absolutely convergent for $\Re(s) > 1$, which guarantees that the sum will converge to the same limit irrespective of order. So
$$ \prod_{p \in \mathbb{P}} \frac1{1- p^{-s}} = \sum_{n \in \mathbb{N^+}}  n^{-s} = \sum_{n = 1}^{\infty}  n^{-s} = \zeta(s) $$

An alternative approach compares the limit with, $\zeta(s)$. Consider,
$$ \prod_{p \in \mathbb{P}} \frac1{1- p^{-s}}- \zeta(s) $$
Then,
$$ \lim_{A \to \infty, K \to \infty} \prod_{p \in \mathbb{P}}^{p \le A} \frac{1- p^{-(K+1)s}}{1- p^{-s}} - \lim_{N \to \infty} \sum_{n = 1}^{N} n^{-s} $$
Or,
$$ \lim_{A \to \infty, K \to \infty} \sum_{n \in Q(A, K)}  n^{-s} - \lim_{N \to \infty} \sum_{n = 1}^{N} n^{-s} $$
So,
$$ \lim_{N \to \infty} \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})}  n^{-s} $$
$s$ may be complex, $ s = u + it $
$$ \lim_{N \to \infty} \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})}  n^{-u} e^{-it \ln(n)}$$
The magnitude is,
$$ \lim_{N \to \infty} | \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})}  n^{-s} | <=  \lim_{N \to \infty}  \sum_{n \in (Q(N, \log_2(N)) - \{1 .. N\})}  n^{-u}$$
$$ <= \lim_{N \to \infty}\sum_{n=N+1}^{\infty} n^{-u} = 0 $$
As $\sum_{n=0}^{\infty} n^{-u}$ converges absolutely for $ u > 1 $
So if $s = u + it \wedge u > 1$,
$$ \prod_{p \in \mathbb{P}} \frac1{1- p^{-s}} = \zeta(s) $$
