= Solution
For $\Re s>1$, the <Euler products> are
$$
\boxed{\zeta(s)=\prod_p(1-p^{-s})^{-1},\qquad
L(s,\chi)=\prod_p(1-\chi(p)p^{-s})^{-1}.}
$$
The <Dirichlet character> is completely multiplicative, and <unique prime factorization> expands each finite product into the terms of the corresponding <Dirichlet series> supported on those primes. Absolute convergence permits passage to all primes. More strongly, the absolutely convergent <Euler product> logarithm gives
$$
L(s,\chi)=\exp\left(\sum_p\sum_{m\ge1}\frac{\chi(p)^m}{m p^{ms}}\right)\ne0,
$$
since the sum of absolute values is bounded by $\sum_{n\ge2}\sum_{m\ge1}n^{-m\Re s}/m<\infty$.
We prove <nonvanishing of a nonprincipal Dirichlet L-function at one>, including real <Dirichlet characters>. If $\chi$ is nonprincipal modulo $q$, periodicity and the given <Orthogonality of Dirichlet characters> bound $A_\chi(x)=\sum_{n\le x}\chi(n)$ by $q$. <Partial summation> gives
$$
L(s,\chi)=s\int_1^\infty A_\chi(x)x^{-s-1}\,dx,
$$
which is holomorphic for $\Re s>0$ by locally uniform convergence, including after differentiation. The <principal Dirichlet character> instead has
$$
L(s,\chi_0)=\zeta(s)\prod_{p\mid q}(1-p^{-s}),
$$
with a simple pole at one.
For a nonreal <Dirichlet character>, consider $P(s)=\prod_{\chi\bmod q}L(s,\chi)$. For real $\sigma>1$, its <Euler product> logarithm is
$$
\log P(\sigma)=\sum_{p\nmid q}\sum_{m\ge1}
\frac{\sum_\chi\chi(p^m)}{m p^{m\sigma}}\ge0.
$$
The given character inversion with $a=1$ makes each inner sum either $\varphi(q)$ or zero, so $P(\sigma)\ge1$. If $L(1,\chi)=0$ for a nonreal <Dirichlet character>, then $L(1,\overline\chi)=0$ as well, since their values at real $s$ are conjugate. These are two distinct factors. All nonprincipal factors are holomorphic at one, and only the principal factor has a simple pole. Their two zeros would force $P(\sigma)\to0$, a contradiction.
For a real nonprincipal <Dirichlet character>, suppose $L(1,\chi)=0$. Then $G(s)=\zeta(s)L(s,\chi)$ is holomorphic throughout $\Re s>0$: the supplied continuation of the <Riemann zeta function> has no other pole there, and the zero cancels the pole at one. On $\Re s>1$ its <Dirichlet series> has coefficients
$$
g_n=\sum_{d\mid n}\chi(d)\ge0.
$$
To check the sign explicitly, for a <prime power> $p^j$ the local sum is $j+1$ if $\chi(p)=1$, is $1$ for even $j$ and $0$ for odd $j$ if $\chi(p)=-1$, and is $1$ if $\chi(p)=0$. Multiplication over <prime factors> proves nonnegativity and $g_{m^2}\ge1$. Also $g_n\le\tau(n)$, so termwise derivatives at $s=2$ are justified by absolute convergence:
$$
(-1)^jG^{(j)}(2)=\sum_n\frac{g_n(\log n)^j}{n^2}\ge0.
$$
The disc $|s-2|<2$ lies in $\Re s>0$. Its <Taylor series> therefore converges at $s=1/2$, and <Tonelli theorem> gives
$$
G(1/2)=\sum_{j\ge0}\frac{(3/2)^j}{j!}\sum_n\frac{g_n(\log n)^j}{n^2}
=\sum_n\frac{g_n}{n^{1/2}}\ge\sum_{m\ge1}\frac1m=\infty.
$$
This contradicts holomorphy at $1/2$. It is the positive-coefficient argument behind the <Landau theorem for a Dirichlet series with nonnegative coefficients>, proved here rather than invoked. Thus $L(1,\chi)\ne0$ for every nonprincipal <Dirichlet character>.
It remains to detect the primes. For real $\sigma>1$, the <Euler product> logarithm satisfies
$$
\log L(\sigma,\chi)=\sum_p\frac{\chi(p)}{p^\sigma}+O(1),
$$
uniformly as $\sigma\downarrow1$: terms with $m\ge2$ are bounded in absolute value by $\sum_{n\ge2}\sum_{m\ge2}n^{-m}/m<\infty$. For a nonprincipal <Dirichlet character>, a <holomorphic logarithm> of $L$ exists near one by the nonvanishing just proved. On a short real interval it differs from the <Euler product> logarithm by a fixed multiple of $2\pi i$, so the prime sum stays bounded. For the <principal Dirichlet character>, the simple pole instead gives
$$
\sum_p\chi_0(p)p^{-\sigma}=\log\frac1{\sigma-1}+O_q(1).
$$
Applying the supplied <Orthogonality of Dirichlet characters> for a reduced residue class $a$ now yields
$$
\sum_{p\equiv a\pmod q}p^{-\sigma}
=\frac1{\varphi(q)}\sum_\chi\overline{\chi(a)}\sum_p\chi(p)p^{-\sigma}
=\frac1{\varphi(q)}\log\frac1{\sigma-1}+O_q(1).
$$
The right side tends to infinity, whereas a finite set of primes would give a bounded left side. Therefore \b[every residue class coprime to $q$ contains infinitely many primes], proving the <Dirichlet theorem on primes in arithmetic progressions>.
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