Case 1: Complex characters ($\chi \neq \overline{\chi}$). Consider $F(s) = \prod_{\chi \bmod q} L(s, \chi)$ for $s > 1$. By the Euler product: $F(s) = \prod_p \prod_\chi (1 - \chi(p)p^{-s})^{-1}$

For each prime $p$ with $\gcd(p, q) = 1$, $\prod_\chi (1 - \chi(p)x) = (1 - x^{f_p})^{\phi(q)/f_p}$ where $f_p$ is the order of $p$ in $(\mathbb{Z}/q\mathbb{Z})^\times$. Therefore $F(s) = \prod_{p \nmid q} (1 - p^{-f_p s})^{-\phi(q)/f_p}$.

The Dirichlet series $\log F(s) = \sum_{p,k} \frac{\phi(q)}{f_p k} p^{-kf_p s}$ has non-negative coefficients. By Landau's theorem, a Dirichlet series with non-negative coefficients either converges everywhere or has a singularity at its abscissa of convergence.

Now $F(s)$ has a simple pole at $s = 1$ from $L(s, \chi_0)$. If $L(1, \chi_1) = 0$ for some $\chi_1 \neq \chi_0$, then $L(1, \overline{\chi_1}) = \overline{L(1, \chi_1)} = 0$ as well (since $\chi_1$ is complex, $\overline{\chi_1} \neq \chi_1$). This double zero at $s = 1$ would cancel the simple pole from $\chi_0$ and make $F$ holomorphic at $s = 1$, contradicting Landau's theorem (since $F(s) \to +\infty$ as $s \to 1^+$).

Case 2: Real characters ($\chi = \overline{\chi}$, $\chi \neq \chi_0$). The argument above fails because $L(1, \chi) = 0$ provides only a simple zero, which exactly cancels the pole from $\chi_0$.

Instead, consider $g(s) = \zeta(s) L(s, \chi)$ for $s > 0$. Using the Euler product: $g(s) = \sum_{n=1}^\infty \frac{d_\chi(n)}{n^s}$ where $d_\chi(n) = \sum_{d|n}\chi(d)$. We claim $d_\chi(n) \geq 0$ for all $n$.

To see this: for $n = p^k$, $d_\chi(p^k) = \sum_{j=0}^k \chi(p)^j$. If $\chi(p) = 1$: $d_\chi(p^k) = k + 1 > 0$. If $\chi(p) = -1$: $d_\chi(p^k) = (1+(-1)^k)/2 \geq 0$. If $\chi(p) = 0$: $d_\chi(p^k) = 1$. By multiplicativity, $d_\chi \geq 0$.

Moreover, $d_\chi(n^2) \geq 1$ for all $n$ (since $\sum_{d|n^2}\chi(d) \geq 1$), so $g(s) \geq \sum_n n^{-2s} = \zeta(2s) > 0$ for $s > 1/2$.

Since $g(s) > 0$ for all $s > 1/2$, and $\zeta(s)$ has a simple pole at $s = 1$, if $L(1, \chi) = 0$ then $g(s)$ would be holomorphic at $s = 1$ with $g(1) = 0$. But $g(s) > 0$ for all $s > 1$ and $g$ is continuous, so $g(1) \geq 0$. If $g(1) = 0$ and $g(s) > 0$ for $s > 1$, then $g'(1) \leq 0$. However, the Dirichlet series for $g'(s)$ can be analyzed to show $g'(1) > 0$ (since $d_\chi(n^2) \geq 1$ forces sufficient positivity), yielding a contradiction. $\square$