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  "body": "\n## Proof\n\nWe expand each factor as a geometric series and bound the error by a tail of the series for $\\zeta(\\sigma)$.\n\n1. Write $s = \\sigma + i t$ with real $\\sigma \u003e 1$ and real $t$. For a real $a \u003e 0$, $a^{-s} = \\exp(-s \\ln a)$.\n2. Two standard facts about the complex exponential are $\\exp(z + w) = \\exp z \\exp w$ and $|\\exp z| = \\exp(\\operatorname{Re} z)$.\n3. With $\\ln(ab) = \\ln a + \\ln b$, they give $(ab)^{-s} = a^{-s} b^{-s}$ for real $a, b \u003e 0$, and $|a^{-s}| = a^{-\\sigma}$.\n4. By the [[p-series test|0ZNSSK8ZK1]], $\\sum_{n \\ge 1} n^{-\\sigma}$ converges, because $\\sigma \u003e 1$. For $N \\ge 1$, let $T_N = \\sum_{n \u003e N} n^{-\\sigma}$. Then $T_N \\to 0$ as $N \\to \\infty$.\n5. The real and imaginary parts of $n^{-s}$ have absolute value at most $|n^{-s}| = n^{-\\sigma}$. By the [[comparison test|04Z5G0THN6]], their series converge absolutely.\n6. By [[absolutely convergent series converge|0Y123X48GA]], both real series converge. So $Z_M = \\sum_{n=1}^{M} n^{-s}$ converges to a limit $\\zeta(s)$ as $M \\to \\infty$.\n7. For a prime $p$, put $r_p = p^{-s}$. Then $|r_p| = p^{-\\sigma} \\le 2^{-\\sigma} \u003c 1/2$. In particular $r_p \\neq 1$.\n8. For $K \\ge 0$, put $G_{p,K} = \\sum_{k=0}^{K} r_p^k$. The sum telescopes: $(1 - r_p) G_{p,K} = 1 - r_p^{K+1}$.\n9. By [[multiplicativity of the modulus|0HYNZE940T]], $|r_p^{K+1}| = |r_p|^{K+1} \u003c 2^{-(K+1)}$, which tends to $0$. So $G_{p,K} \\to 1/(1 - r_p)$ as $K \\to \\infty$.\n10. Fix an integer $N \\ge 2$. Let $q_1 \u003c \\dots \u003c q_m$ be the primes at most $N$. Put $P_N = \\prod_{j=1}^{m} 1/(1 - r_{q_j})$ and $F_{N,K} = \\prod_{j=1}^{m} G_{q_j,K}$.\n11. A product of finitely many convergent complex sequences converges to the product of the limits. This follows from the [[real case|0734RSVF2V]] applied to real and imaginary parts. So $F_{N,K} \\to P_N$ as $K \\to \\infty$.\n12. Multiplying out, $F_{N,K}$ is the sum, over all tuples $(k_1, \\dots, k_m)$ with $0 \\le k_j \\le K$, of $\\prod_j r_{q_j}^{k_j}$. By step 3, this term is $n^{-s}$ with $n = q_1^{k_1} \\cdots q_m^{k_m}$.\n13. By the [[fundamental theorem of arithmetic|084HH1NDCB]], distinct tuples give distinct integers $n$. Let $S_{N,K}$ be the set of these integers. Then $F_{N,K} = \\sum_{n \\in S_{N,K}} n^{-s}$.\n14. Let $K \\ge N$. Each integer $n$ with $1 \\le n \\le N$ lies in $S_{N,K}$. Indeed, by the [[existence of prime factorization|020K0X7HVR]], $n$ is a product of primes at most $n \\le N$.\n15. In that factorization, a prime $q$ with exponent $e$ satisfies $2^e \\le q^e \\le n \\le N \\le K \u003c 2^K$. So $e \\le K$, and $n \\in S_{N,K}$.\n16. Let $M$ be at least the largest element of $S_{N,K}$. Then $Z_M - F_{N,K}$ is the sum of $n^{-s}$ over the $n \\le M$ with $n \\notin S_{N,K}$. By step 14, all these $n$ exceed $N$.\n17. By the [[triangle inequality|0FJZMBX0ZZ]] for the modulus, $|Z_M - F_{N,K}| \\le \\sum_{N \u003c n \\le M} n^{-\\sigma} \\le T_N$.\n18. Let $M \\to \\infty$. Since $Z_M \\to \\zeta(s)$ and the modulus is continuous, $|\\zeta(s) - F_{N,K}| \\le T_N$ for every $K \\ge N$.\n19. Let $K \\to \\infty$. By step 11 and continuity of the modulus, $|\\zeta(s) - P_N| \\le T_N$.\n20. Let $N \\to \\infty$. By step 4, $P_N \\to \\zeta(s)$. The numbers $P_N$ are the partial products of $\\prod_p 1/(1 - p^{-s})$ over the primes in increasing order. So the product equals $\\zeta(s)$.\n21. For absolute convergence, write $1/(1 - r_p) = 1 + a_p$ with $a_p = r_p/(1 - r_p)$. We show that $\\sum_p |a_p|$ converges.\n22. By the triangle inequality, $|1 - r_p| \\ge 1 - |r_p| \u003e 1/2$. So $|a_p| \u003c 2 |r_p| = 2 p^{-\\sigma}$.\n23. The partial sums of $\\sum_p |a_p|$ increase, and they are at most $2 \\sum_{n \\ge 1} n^{-\\sigma}$. By the [[monotone convergence theorem|0FM6N4R0VB]], $\\sum_p |a_p|$ converges.\n24. So the product $\\prod_p 1/(1 - p^{-s})$ converges absolutely, and its value is $\\zeta(s)$.\n\n## Idea\n\nEach factor $1/(1 - p^{-s})$ is a geometric series $\\sum_k p^{-ks}$. Multiplying these series over the primes up to $N$ gives $\\sum n^{-s}$ over the integers built from those primes, each exactly once by unique factorization. Every $n \\le N$ appears, so the error is at most the tail $\\sum_{n \u003e N} n^{-\\sigma}$, which tends to $0$.\n",
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