[proofplan] We use the factorisation $\zeta_{\mathbb{Q}(\omega_q)}(s) = \prod_{i=1}^{\varphi(q)} L(\chi_i, s)$ from [Theorem 1623](/theorems/1623). At $s = 1$, the left side has a simple pole (order of zero $= -1$) by the analytic class number formula. The factor $L(\chi_1, s) = \zeta_\mathbb{Q}(s) \cdot (\text{finite Euler product})$ has a simple pole at $s = 1$. The other factors $L(\chi_i, s)$ for $i \geq 2$ are each holomorphic at $s = 1$ by [Holomorphicity of Non-Trivial $L$-Functions](/theorems/1620), so their orders of vanishing are non-negative integers. Summing orders of vanishing across all factors: left side gives $-1$; right side gives $-1 + \sum_{i \geq 2} \operatorname{ord}_{s=1} L(\chi_i, s)$. Equating these and using non-negativity of each summand forces each $\operatorname{ord}_{s=1} L(\chi_i, s) = 0$, i.e., $L(\chi_i, 1) \neq 0$ for $i \geq 2$. [/proofplan] [step:Apply the factorisation and read off orders of vanishing at $s = 1$] Let $\chi$ be a non-trivial Dirichlet character of modulus $q$. Then $\chi$ appears as one of the $\varphi(q)$ characters of modulus $q$, i.e., $\chi = \chi_{i_0}$ for some $i_0 \geq 2$ (with $\chi_1$ the trivial character, and $\chi_2, \ldots, \chi_{\varphi(q)}$ the non-trivial ones). By [Factorisation of the Cyclotomic Zeta Function](/theorems/1623): the hypothesis is that $L = \mathbb{Q}(\omega_q)$, which is the $q$-th cyclotomic field, and $\chi_1, \ldots, \chi_{\varphi(q)}$ are the Dirichlet characters of modulus $q$. The conclusion: \begin{align*} \zeta_{\mathbb{Q}(\omega_q)}(s) &= \prod_{i = 1}^{\varphi(q)} L(\chi_i, s). \end{align*} Taking the order of vanishing at $s = 1$ on both sides, using additivity of the order of vanishing under multiplication of meromorphic functions: \begin{align*} \operatorname{ord}_{s = 1} \zeta_{\mathbb{Q}(\omega_q)}(s) &= \sum_{i = 1}^{\varphi(q)} \operatorname{ord}_{s = 1} L(\chi_i, s). \end{align*} [guided] **Meromorphic order of vanishing.** For a meromorphic function $f$ defined near $s_0 \in \mathbb{C}$, the order $\operatorname{ord}_{s_0} f$ is the unique integer $k$ such that $f(s) = (s - s_0)^k g(s)$ for a holomorphic $g$ with $g(s_0) \neq 0$. If $f$ is holomorphic and non-zero at $s_0$, $\operatorname{ord}_{s_0} f = 0$. A simple zero has order $1$; a simple pole has order $-1$. **Additivity under products.** If $f = \prod_i f_i$ is a product of finitely many meromorphic functions near $s_0$, with each $f_i = (s - s_0)^{k_i} g_i(s)$, then \begin{align*} f(s) = \prod_i (s - s_0)^{k_i} g_i(s) = (s - s_0)^{\sum_i k_i} \prod_i g_i(s), \end{align*} and $\operatorname{ord}_{s_0} f = \sum_i k_i = \sum_i \operatorname{ord}_{s_0} f_i$. **Applying to our factorisation.** The Dedekind zeta function $\zeta_L$ and each $L(\chi_i, s)$ are meromorphic on a neighbourhood of $s = 1$ (actually in a larger domain, but this is all we need). Taking orders of vanishing: \begin{align*} \operatorname{ord}_{s=1}\zeta_L = \sum_{i = 1}^{\varphi(q)} \operatorname{ord}_{s=1} L(\chi_i, s). \end{align*} This is the key equation. We now compute each side. [/guided] [/step] [step:Compute $\operatorname{ord}_{s = 1} \zeta_{\mathbb{Q}(\omega_q)} = -1$] By the analytic class number formula for the Dedekind zeta function of $\mathbb{Q}(\omega_q)$: $\zeta_{\mathbb{Q}(\omega_q)}(s)$ has a **simple pole** at $s = 1$ with non-zero residue \begin{align*} \operatorname{Res}_{s = 1} \zeta_{\mathbb{Q}(\omega_q)}(s) &= \frac{2^r (2\pi)^s h_L R_L}{w_L |D_L|^{1/2}} > 0, \end{align*} where $(r, s)$ is the signature of $\mathbb{Q}(\omega_q)$, $h_L$ is its class number, $R_L$ its regulator, $w_L$ the number of roots of unity, and $D_L$ the discriminant — all positive real numbers. Hence \begin{align*} \operatorname{ord}_{s = 1} \zeta_{\mathbb{Q}(\omega_q)}(s) &= -1. \end{align*} [guided] **Analytic class number formula.** For any number field $L$, the Dedekind zeta function $\zeta_L$ is a meromorphic function on $\mathbb{C}$ whose only pole is a simple pole at $s = 1$, with residue given by the analytic class number formula \begin{align*} \operatorname{Res}_{s=1}\zeta_L(s) = \frac{2^r (2\pi)^s h_L R_L}{w_L \sqrt{|D_L|}}. \end{align*} In our setting $L = \mathbb{Q}(\omega_q)$: all of $r, s$ (signature components), $h_L, R_L, w_L, |D_L|$ are positive real numbers, so the residue is positive and non-zero. **Order of vanishing.** A simple pole at $s = 1$ means $\zeta_L(s) = (s - 1)^{-1} g(s)$ near $s = 1$ for some $g$ holomorphic with $g(1) \neq 0$. In the conventions for order of vanishing, this is $\operatorname{ord}_{s=1}\zeta_L = -1$. [/guided] [/step] [step:Compute $\operatorname{ord}_{s = 1} L(\chi_1, s) = -1$] The trivial character $\chi_1$ (the Dirichlet character of modulus $q$ taking value $1$ on all units) has the Euler product expansion \begin{align*} L(\chi_1, s) &= \prod_{p \nmid q} \bigl(1 - p^{-s}\bigr)^{-1} = \zeta_\mathbb{Q}(s) \prod_{p \mid q} \bigl(1 - p^{-s}\bigr). \end{align*} The second factor, $\prod_{p \mid q} (1 - p^{-s})$, is a finite product of entire functions, hence entire. At $s = 1$, $\prod_{p \mid q}(1 - p^{-1}) = \varphi(q)/q > 0$, so this factor is non-zero at $s = 1$. $\zeta_\mathbb{Q}(s)$ has a simple pole at $s = 1$ with residue $1$ (Step 2 of [Non-Vanishing for Quadratic Characters](/theorems/1621)). Therefore \begin{align*} \operatorname{ord}_{s = 1} L(\chi_1, s) &= \operatorname{ord}_{s = 1} \zeta_\mathbb{Q}(s) + \operatorname{ord}_{s = 1} \prod_{p \mid q} \bigl(1 - p^{-s}\bigr) = -1 + 0 = -1. \end{align*} [guided] **Trivial character.** The trivial Dirichlet character of modulus $q$ is \begin{align*} \chi_1(n) = \begin{cases} 1 & \gcd(n, q) = 1, \\ 0 & \gcd(n, q) > 1. \end{cases} \end{align*} Its $L$-function, via the Euler product formula for Dirichlet $L$-series, \begin{align*} L(\chi_1, s) = \prod_p (1 - \chi_1(p) p^{-s})^{-1} = \prod_{p \nmid q} (1 - p^{-s})^{-1}, \end{align*} since $\chi_1(p) = 1$ for $p \nmid q$ and $\chi_1(p) = 0$ for $p \mid q$ (which contributes a factor of $1$, i.e., nothing). **Relating to $\zeta_\mathbb{Q}$.** Multiplying and dividing by the missing Euler factors at $p \mid q$: \begin{align*} L(\chi_1, s) = \frac{\prod_p (1 - p^{-s})^{-1}}{\prod_{p \mid q}(1 - p^{-s})^{-1}} = \zeta_\mathbb{Q}(s) \cdot \prod_{p \mid q}(1 - p^{-s}). \end{align*} The finite product $\prod_{p \mid q}(1 - p^{-s})$ consists of finitely many entire functions, each non-zero at $s = 1$ ($1 - p^{-1} > 0$ for $p \geq 2$), so the product is entire and non-zero at $s = 1$. **Order of vanishing of $L(\chi_1, s)$ at $s = 1$.** By additivity: \begin{align*} \operatorname{ord}_{s=1}L(\chi_1, s) = \operatorname{ord}_{s=1}\zeta_\mathbb{Q}(s) + \operatorname{ord}_{s=1}\prod_{p \mid q}(1 - p^{-s}) = -1 + 0 = -1. \end{align*} $L(\chi_1, s)$ has a simple pole at $s = 1$ just like $\zeta_\mathbb{Q}$. [/guided] [/step] [step:Use non-negativity of orders for $i \geq 2$ to conclude $L(\chi_i, 1) \neq 0$] For each $i \geq 2$, $\chi_i$ is a non-trivial Dirichlet character of modulus $q$. By [Holomorphicity of Non-Trivial $L$-Functions](/theorems/1620): the hypothesis is that $\chi_i$ is non-trivial, which holds; the conclusion is that $L(\chi_i, s)$ is holomorphic on $\{\operatorname{Re}(s) > 0\}$, in particular at $s = 1$. Holomorphicity at $s = 1$ means \begin{align*} \operatorname{ord}_{s = 1} L(\chi_i, s) &\geq 0 \qquad \text{for each } i \geq 2. \end{align*} Substituting into the order-summation identity from Step 1: \begin{align*} -1 &= \operatorname{ord}_{s = 1}\zeta_{\mathbb{Q}(\omega_q)}(s) = \sum_{i = 1}^{\varphi(q)} \operatorname{ord}_{s = 1} L(\chi_i, s) = \operatorname{ord}_{s = 1} L(\chi_1, s) + \sum_{i = 2}^{\varphi(q)} \operatorname{ord}_{s = 1} L(\chi_i, s) \\ &= -1 + \sum_{i = 2}^{\varphi(q)} \operatorname{ord}_{s = 1} L(\chi_i, s). \end{align*} Cancelling the $-1$ on both sides: \begin{align*} 0 &= \sum_{i = 2}^{\varphi(q)} \operatorname{ord}_{s = 1} L(\chi_i, s). \end{align*} Each summand is a non-negative integer (from the holomorphicity bound). A sum of non-negative integers is zero if and only if every summand is zero: \begin{align*} \operatorname{ord}_{s = 1} L(\chi_i, s) &= 0 \qquad \text{for every } i = 2, \ldots, \varphi(q). \end{align*} In other words, $L(\chi_i, s)$ is holomorphic and non-zero at $s = 1$, i.e., $L(\chi_i, 1) \neq 0$. Since $\chi$ was an arbitrary non-trivial Dirichlet character (matching some $\chi_i$ with $i \geq 2$), this proves $L(\chi, 1) \neq 0$ for every non-trivial Dirichlet character $\chi$. [guided] **Setup.** From Step 1 we have the order identity \begin{align*} \operatorname{ord}_{s = 1}\zeta_{\mathbb{Q}(\omega_q)} = \sum_{i = 1}^{\varphi(q)} \operatorname{ord}_{s = 1} L(\chi_i, s). \end{align*} Step 2 gave the left side as $-1$; Step 3 gave $\operatorname{ord}_{s=1} L(\chi_1, s) = -1$. **Holomorphy for non-trivial characters.** By [Theorem 1620](/theorems/1620), each non-trivial $L(\chi_i, s)$ (for $i \geq 2$) is holomorphic on $\{\operatorname{Re}(s) > 0\}$, and in particular at $s = 1$. Holomorphy means $\operatorname{ord}_{s=1} L(\chi_i, s) \geq 0$ (integer-valued; zero if non-vanishing, positive if a zero of some order). **Counting orders.** The identity becomes \begin{align*} -1 = -1 + \sum_{i \geq 2} \operatorname{ord}_{s=1} L(\chi_i, s), \end{align*} so $\sum_{i \geq 2} \operatorname{ord}_{s=1} L(\chi_i, s) = 0$. **Crucial observation.** Each term is $\geq 0$, and their sum is $0$. A finite sum of non-negative integers equals zero if and only if each term is zero. Hence $\operatorname{ord}_{s=1} L(\chi_i, s) = 0$ for every $i \geq 2$. **Translation.** Order zero means holomorphic and non-vanishing. Therefore $L(\chi_i, 1)$ is a well-defined non-zero complex number for every non-trivial character $\chi_i$. **Why this is non-trivial.** Holomorphicity alone (Theorem 1620) does not preclude $L(\chi, 1) = 0$; only the factorisation through $\zeta_L$ enforces that a zero of any $L(\chi_i, s)$ at $s = 1$ would be visible in the pole structure of $\zeta_L$, which has only the contribution from $\zeta_\mathbb{Q} = L(\chi_1, \cdot)$. Any additional zero would require an additional pole elsewhere, which is impossible. [/guided] [/step]