For $s > 1,$ we have $\zeta(s) = \prod \left( \dfrac{1}{1-p^{-s}} \right)$ 
For $s > 1,$ we have $\zeta(s) = \displaystyle \prod \left(\dfrac{1}{1-p^{-s}} \right).$

Fix a positive integer $y,$ let $p_1, \ldots, p_n$ be primes and define $N_y := \{n \in \mathbb{N}: \text{ If } p_i \leqslant y, \text{ then } p_i \mid n \}.$ Fix an $\epsilon > 0.$ Note that for every prime $p \leqslant y,$ one has $$\dfrac{1}{1-p^{-s}} = \lim_{k \to \infty} (1 + p^{-s} + \ldots + p^{-ks}).$$ Since there are finitely many primes $p \leqslant y,$ we can choose a $k$ sufficiently large so that 
$$(\ast) \hspace{1mm} \left| \prod_{p \leqslant y} \left(  \dfrac{1}{1-p^{-s}} \right) - \prod_{p \leqslant y} \left(1 + p^{-s} + \cdots + p^{-ks} \right) \right| < \epsilon $$ for each prime $p \leqslant y.$ Expanding out the second product, we obtain the sum of $n^{-s}$ for all $n$ in the set $$N_{y,k} = \{n \in \mathbb{Z}: n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_r^{\alpha_r} \text{ with every } p_i \leqslant y \text{ and every } \alpha_i \leqslant k \}.$$ Since the $N_{y,k}$ for increasing $k$ are nested of integers whose union is $N_y,$ we can take $k$ sufficiently large so that $$ (\star) \hspace{1mm} \left| \sum_{n \in N_y} \dfrac{1}{n^s} - \sum_{n \in N_{y,k}} \dfrac{1}{n} \right| < \epsilon.$$
Question: How do I combine $(\star)$ and $(\ast) $ to obtain $$\left| \prod_{p \leqslant y} \left( \dfrac{1}{1-p^{-s}} \right) - \sum_{n \in N_y} \dfrac{1}{n^s} \right| < 2 \epsilon?$$
 A: Lemma: The product $\displaystyle \prod\limits_{n=1}^{\infty}(1+a_n)$ converges absolutely iff $\displaystyle \sum\limits_{n=1}^{\infty} |a_n| < \infty$.
Proof: The infinite product $\displaystyle \prod\limits_{n=1}^{\infty}(1+a_n)$ converges absolutely by definition if the sequence of partial product of terms $\{P_N^{*}\}:= \prod\limits_{n=1}^{N}(1+|a_n|)$ converges. Given, that $\displaystyle \sum\limits_{n=1}^{\infty} |a_n| < \infty$ we have $P_N^{*} \le \exp({\sum\limits_{n=1}^{\infty} |a_n|})$ where, we used the inequality $1+x \le e^x$ for, $x \ge 0$. Conversely, if the product converges then $\displaystyle P_n^{*} \ge 1+\sum\limits_{n=1}^{\infty} |a_n| \implies \sum\limits_{n=1}^{\infty} |a_n| < \infty$ by monotone convergence theorem.
Now, the convergence of the nfinite product $\displaystyle \prod\limits_{p}(1-p^{-s})$ follows form the lemma, since the product is bounded
 by the sum $\displaystyle \sum\limits_{p} \left|\sum\limits_{m=1}^{\infty}\frac{1}{p^{ms}}\right| \le \sum\limits_{n=1}^{\infty} \frac{1}{n^s}$
Therefore, it remains to show that the limit $\displaystyle \lim\limits_{N \to \infty} \prod\limits_{p \le N}\left(1+\sum\limits_{m=1}^{\infty}\frac{1}{p^{ms}}\right)$ equals $\zeta(s)$.
Let, $p_1,p_2,\cdots,p_k$ be the primes less than or equal to $N$. 
Now, $\displaystyle \prod\limits_{p \le N}\left(1+\sum\limits_{m=1}^{\infty}\frac{1}{p^{ms}}\right) = \sum\limits_{m_1 = 0}^{\infty}\cdots \sum\limits_{m_k = 0}^{\infty} \dfrac{1}{p_1^{m_1s}\cdots p_k^{m_ks}}$
The integers appearing in the denominator of these summation are unique (by unique prime factorization of integers) and are only composed of primes less than or equal to $N$. Thus, if $\displaystyle P_{N} = \{n \in \mathbb{N} | \textrm{ s.t. } p|n \implies p \le N \}$ we can write the sum $\displaystyle \sum\limits_{m_1 = 0}^{\infty}\cdots \sum\limits_{m_k = 0}^{\infty} \dfrac{1}{p_1^{m_1s}\cdots p_k^{m_ks}} = \sum\limits_{n \in P_N} \frac{1}{n^s}$
Now, $P_N$ contains each integer $\le N$, hence, $\displaystyle \left|\zeta(s) - \sum\limits_{m_1 = 0}^{\infty}\cdots \sum\limits_{m_k = 0}^{\infty} \dfrac{1}{p_1^{m_1s}\cdots p_k^{m_ks}}\right| = \left|\sum\limits_{n \notin P_N}\frac{1}{n^s}\right| < \left|\sum\limits_{n > N} \frac{1}{n^s}\right| \to 0$ as $n \to \infty$.
Hence, $$\displaystyle \lim\limits_{N \to \infty} \prod\limits_{p \le N}\left(1+\sum\limits_{m=1}^{\infty}\frac{1}{p^{ms}}\right) = \prod\limits_{p} \left(1+\sum\limits_{m=1}^{\infty}\frac{1}{p^{ms}}\right) = \prod\limits_{p}(1-p^{-s})^{-1} = \zeta(s)$$ for $\mathfrak{Re}(s) > 1$.
