[proofplan] We prove the two assertions in sequence. For (i), we show that the partial product $\prod_{p \leq x}(1 - p^{-s})^{-1}$ converges to $\zeta(s)$ as $x \to \infty$ by identifying the difference as the tail sum over integers divisible by some prime exceeding $x$; this tail vanishes because $\sum n^{-\sigma}$ converges for $\sigma = \operatorname{Re}(s) > 1$. For (ii), we leverage (i) to write $\zeta(s)$ as a quotient of a convergent nonzero partial Euler product and a quantity bounded away from zero for large $x$, forcing $\zeta(s) \neq 0$. The unique-factorisation identity expanding a finite Euler product into a sum over integers with bounded prime factors is the structural engine of both parts. [/proofplan] [step:Expand the finite Euler product over primes up to $x$ using geometric series and unique factorisation] Fix $s = \sigma + it \in \mathbb{C}$ with $\sigma > 1$, and fix $x \geq 2$. For each prime $p$, the geometric series identity gives \begin{align*} \left(1 - \frac{1}{p^s}\right)^{-1} = \sum_{k=0}^\infty \frac{1}{p^{ks}}, \end{align*} with absolute convergence because $|p^{-s}| = p^{-\sigma} < 1$. Multiplying these series over all primes $p \leq x$ and using absolute convergence to rearrange, \begin{align*} \prod_{p \leq x} \left(1 - \frac{1}{p^s}\right)^{-1} = \sum_{n \in S_x} \frac{1}{n^s}, \end{align*} where \begin{align*} S_x := \{n \in \mathbb{N} : \text{every prime factor of } n \text{ is } \leq x\}. \end{align*} By unique prime factorisation, each $n \in S_x$ corresponds bijectively to a tuple of exponents $(k_p)_{p \leq x}$ with $n = \prod_{p \leq x} p^{k_p}$, so each $n \in S_x$ contributes the term $n^{-s}$ exactly once. [guided] We want to rewrite the finite Euler product as a Dirichlet series. The tool is the geometric series: for $|z| < 1$, $(1-z)^{-1} = \sum_{k \geq 0} z^k$. Applied with $z = p^{-s}$, convergence requires $|p^{-s}| = p^{-\sigma} < 1$, which holds because $p \geq 2$ and $\sigma > 1$. Thus \begin{align*} \left(1 - \frac{1}{p^s}\right)^{-1} = 1 + \frac{1}{p^s} + \frac{1}{p^{2s}} + \cdots = \sum_{k \geq 0} \frac{1}{p^{ks}}. \end{align*} Now we multiply these series over $p \leq x$, which is a finite product of absolutely convergent series. Absolute convergence allows us to expand and regroup without worrying about order: \begin{align*} \prod_{p \leq x} \left(1 - \frac{1}{p^s}\right)^{-1} = \prod_{p \leq x} \sum_{k_p \geq 0} \frac{1}{p^{k_p s}} = \sum_{(k_p)_{p \leq x}} \prod_{p \leq x} \frac{1}{p^{k_p s}} = \sum_{(k_p)_{p \leq x}} \frac{1}{\left(\prod_{p \leq x} p^{k_p}\right)^s}. \end{align*} Each tuple $(k_p)_{p \leq x}$ of non-negative integers determines a unique positive integer $n = \prod_{p \leq x} p^{k_p}$, and by unique factorisation the map $(k_p) \mapsto n$ is a bijection onto \begin{align*} S_x := \{n \in \mathbb{N} : \text{every prime factor of } n \text{ is } \leq x\}. \end{align*} Therefore \begin{align*} \prod_{p \leq x} \left(1 - \frac{1}{p^s}\right)^{-1} = \sum_{n \in S_x} \frac{1}{n^s}. \end{align*} This is the clean form we will compare against $\zeta(s) = \sum_{n \geq 1} n^{-s}$ (absolute convergence for $\sigma > 1$ is the content of the [Convergence of the Zeta Series](/theorems/1746)). [/guided] [/step] [step:Control the tail $\sum_{n \in \eta_x} n^{-s}$ using absolute convergence on $\sigma > 1$] Let $\eta_x := \mathbb{N} \setminus S_x = \{n \geq 1 : n \text{ has at least one prime factor } > x\}$. Subtracting the identity from Step 1 from $\zeta(s) = \sum_{n \geq 1} n^{-s}$, \begin{align*} \zeta(s) - \prod_{p \leq x} \left(1 - \frac{1}{p^s}\right)^{-1} = \sum_{n \in \eta_x} \frac{1}{n^s}. \end{align*} Every $n \in \eta_x$ has a prime factor exceeding $x$, so $n > x$, giving $\eta_x \subseteq \{n > x\}$. Using $|n^{-s}| = n^{-\sigma}$ and the triangle inequality, \begin{align*} \left| \zeta(s) - \prod_{p \leq x} \left(1 - \frac{1}{p^s}\right)^{-1} \right| \leq \sum_{n \in \eta_x} n^{-\sigma} \leq \sum_{n > x} n^{-\sigma}. \end{align*} Since $\sigma > 1$, the series $\sum_{n \geq 1} n^{-\sigma}$ converges by the [Convergence of the Zeta Series](/theorems/1746), so its tail satisfies $\sum_{n > x} n^{-\sigma} \to 0$ as $x \to \infty$. Hence \begin{align*} \prod_{p \leq x} \left(1 - \frac{1}{p^s}\right)^{-1} \xrightarrow[x \to \infty]{} \zeta(s), \end{align*} which is precisely the statement of the Euler product identity (i). [guided] We have two quantities: \begin{align*} \zeta(s) = \sum_{n \geq 1} n^{-s}, \qquad \prod_{p \leq x}(1-p^{-s})^{-1} = \sum_{n \in S_x} n^{-s}. \end{align*} Subtracting, the difference is the sum over $\mathbb{N} \setminus S_x = \eta_x$: \begin{align*} \zeta(s) - \prod_{p \leq x}\left(1 - \frac{1}{p^s}\right)^{-1} = \sum_{n \in \eta_x} \frac{1}{n^s}. \end{align*} What is $\eta_x$? It is the set of $n$ whose prime factorisation contains at least one prime $p > x$. In particular, any $n \in \eta_x$ is divisible by some $p > x$, so $n \geq p > x$. This embedding $\eta_x \subseteq \{n > x\}$ is the key observation — it allows us to bound the tail by the Dirichlet tail. Taking absolute values, $|n^{-s}| = |e^{-s \log n}| = e^{-\sigma \log n} = n^{-\sigma}$, so \begin{align*} \left| \sum_{n \in \eta_x} n^{-s} \right| \leq \sum_{n \in \eta_x} |n^{-s}| = \sum_{n \in \eta_x} n^{-\sigma} \leq \sum_{n > x} n^{-\sigma}. \end{align*} The last inequality uses $\eta_x \subseteq \{n > x\}$ and positivity of all terms. The right-hand side is the tail of the convergent series $\sum n^{-\sigma}$ (from the [Convergence of the Zeta Series](/theorems/1746), valid because $\sigma > 1$), so it tends to zero as $x \to \infty$. Conclusion: \begin{align*} \prod_{p \leq x} \left(1 - \frac{1}{p^s}\right)^{-1} \xrightarrow[x \to \infty]{} \zeta(s). \end{align*} [/guided] [/step] [step:Invert the Euler product relation to extract $\zeta(s)$ as a nonzero limit] For (ii), fix $s$ with $\sigma = \operatorname{Re}(s) > 1$. Using (i) and multiplying both sides of the identity $\prod_{p \leq x}(1-p^{-s})^{-1} \cdot \prod_{p \leq x}(1-p^{-s}) = 1$ by $\zeta(s)$, \begin{align*} \prod_{p \leq x} \left(1 - \frac{1}{p^s}\right) \cdot \zeta(s) = \frac{\zeta(s)}{\prod_{p \leq x}(1-p^{-s})^{-1}}. \end{align*} Now $\prod_{p \leq x}(1-p^{-s})^{-1} = \sum_{n \in S_x} n^{-s}$ by Step 1, so dividing $\zeta(s) = \sum_{n \in S_x} n^{-s} + \sum_{n \in \eta_x} n^{-s}$ and manipulating gives the cleaner formulation: expand the finite product $\prod_{p \leq x}(1-p^{-s})^{-1} \cdot \zeta(s)$ by the same unique-factorisation argument as Step 1 but now keeping the factor $\zeta(s)$ as a Dirichlet series. We obtain \begin{align*} \prod_{p \leq x} \left(1 - \frac{1}{p^s}\right) \cdot \zeta(s) = 1 + \sum_{\substack{n \geq 2 \\ p \mid n \implies p > x}} \frac{1}{n^s}. \end{align*} [guided] We want to show $\zeta(s) \neq 0$ for $\sigma > 1$. The strategy is to pin down $\zeta(s)$ times a finite product and show the result is close to $1$, forcing $\zeta(s)$ to be nonzero. The identity $\prod_{p \leq x}(1 - p^{-s}) \cdot \zeta(s) = ?$ should have a clean expansion. We compute directly. Write \begin{align*} \prod_{p \leq x}\left(1 - \frac{1}{p^s}\right) \cdot \zeta(s) = \prod_{p \leq x}\left(1 - \frac{1}{p^s}\right) \cdot \sum_{n \geq 1} \frac{1}{n^s}. \end{align*} Expanding $\prod_{p \leq x}(1 - p^{-s})$ as a finite sum over subsets $T \subseteq \{p : p \leq x\}$, \begin{align*} \prod_{p \leq x}\left(1 - \frac{1}{p^s}\right) = \sum_{T \subseteq \{p \leq x\}} (-1)^{|T|} \prod_{p \in T} \frac{1}{p^s} = \sum_{T} \frac{(-1)^{|T|}}{(\prod_{p \in T} p)^s}. \end{align*} Multiplying by $\zeta(s) = \sum_m m^{-s}$ and using absolute convergence, \begin{align*} \prod_{p \leq x}\left(1 - \frac{1}{p^s}\right) \cdot \zeta(s) = \sum_{m \geq 1} \sum_T \frac{(-1)^{|T|}}{(m \prod_{p \in T} p)^s}. \end{align*} For a fixed positive integer $N = m \prod_{p \in T} p$, collecting terms gives a coefficient \begin{align*} c_N = \sum_{\substack{T \subseteq \{p \leq x\} \\ \prod_{p \in T} p \mid N}} (-1)^{|T|} = \prod_{p \leq x}\left(1 - [p \mid N]\right), \end{align*} where $[p \mid N] \in \{0,1\}$ is the indicator that $p \mid N$. This factorises because $T$ ranges independently over subsets, and for each $p$ the contribution is $1$ (if $p \nmid N$, take $p \notin T$ only) or $1 + (-1) = 0$ (if $p \mid N$, both $p \in T$ and $p \notin T$ are allowed). Hence $c_N = 0$ unless no prime $p \leq x$ divides $N$, and $c_N = 1$ if every prime factor of $N$ exceeds $x$ (or $N = 1$, where the empty product leaves $c_1 = 1$). Thus \begin{align*} \prod_{p \leq x}\left(1 - \frac{1}{p^s}\right) \cdot \zeta(s) = 1 + \sum_{\substack{N \geq 2 \\ p \mid N \implies p > x}} \frac{1}{N^s}. \end{align*} This is the desired expansion: $\zeta(s)$ times a finite product equals $1$ plus a tail of terms supported on integers all of whose prime factors exceed $x$. [/guided] [/step] [step:Bound the tail modulus below $1$ for large $x$ to force $\zeta(s) \neq 0$] Every index $N \geq 2$ in the tail sum from Step 3 has all prime factors $> x$, so in particular $N > x$. Therefore \begin{align*} \left| \sum_{\substack{N \geq 2 \\ p \mid N \implies p > x}} \frac{1}{N^s} \right| \leq \sum_{N > x} N^{-\sigma}, \end{align*} and the right-hand side tends to $0$ as $x \to \infty$ by the [Convergence of the Zeta Series](/theorems/1746) applied at exponent $\sigma > 1$. Choose $x_0$ large enough that $\sum_{N > x_0} N^{-\sigma} < \tfrac{1}{2}$. Then for all $x \geq x_0$, \begin{align*} \left| \prod_{p \leq x} \left(1 - \frac{1}{p^s}\right) \cdot \zeta(s) \right| \geq 1 - \frac{1}{2} = \frac{1}{2} > 0. \end{align*} The finite product $\prod_{p \leq x}(1 - p^{-s})$ is a product of finitely many nonzero complex numbers (each factor is nonzero because $|p^{-s}| = p^{-\sigma} < 1$ forces $1 - p^{-s} \neq 0$), so it is itself nonzero. The product of a nonzero complex number and $\zeta(s)$ has nonzero modulus, forcing $\zeta(s) \neq 0$. Combining Steps 1 through 4 gives both assertions: the partial Euler product converges to $\zeta(s)$, and $\zeta(s)$ is nonzero throughout the half-plane $\sigma > 1$. [/step]