[proofplan]
We first express $L(s,\chi)$ as an absolutely convergent Euler product in the half-plane $\operatorname{Re}(s)>1$. We then take logarithms of the local factors and justify differentiating the resulting normally convergent series term by term. The differentiated prime-power expansion is finally reindexed by integers, using complete multiplicativity of $\chi$ and the definition of the von Mangoldt function.
[/proofplan]
[step:Establish absolute convergence and the Euler product in the half-plane]
Fix $s \in \mathbb{C}$ and write $\sigma := \operatorname{Re}(s)$. Since $|\chi(n)| \leq 1$ for every $n \in \mathbb{N}$, the Dirichlet series defining $L(s,\chi)$ is absolutely convergent for $\sigma>1$, because
\begin{align*}
\sum_{n=1}^{\infty} \left|\frac{\chi(n)}{n^s}\right|
\leq
\sum_{n=1}^{\infty} \frac{1}{n^\sigma}
<\infty.
\end{align*}
For a finite set $P$ of primes, define
\begin{align*}
E_P(\cdot): \{s \in \mathbb{C} : \operatorname{Re}(s)>1\} &\to \mathbb{C} \\
s &\mapsto \prod_{p \in P} \left(1-\chi(p)p^{-s}\right)^{-1}.
\end{align*}
Since $|\chi(p)p^{-s}| \leq p^{-\sigma}<1$, each factor has the absolutely convergent geometric expansion
\begin{align*}
\left(1-\chi(p)p^{-s}\right)^{-1}
=
\sum_{k=0}^{\infty}\chi(p)^k p^{-ks}.
\end{align*}
Multiplying finitely many absolutely convergent geometric series gives
\begin{align*}
E_P(s)
=
\sum_{\substack{n \geq 1\\ \text{all prime divisors of } n \text{ lie in } P}}
\frac{\chi(n)}{n^s},
\end{align*}
because every integer supported on $P$ has a unique factorisation $n=\prod_{p \in P}p^{k_p}$ and $\chi$ is completely multiplicative. Letting $P$ increase through the finite sets of primes, the right-hand side converges absolutely to $\sum_{n=1}^{\infty}\chi(n)n^{-s}$. Hence
\begin{align*}
L(s,\chi)
=
\prod_p \left(1-\chi(p)p^{-s}\right)^{-1}
\end{align*}
for every $\sigma>1$.
[guided]
The first point is that no conditional convergence is being used. For $\sigma=\operatorname{Re}(s)>1$, the estimate $|\chi(n)|\leq 1$ gives
\begin{align*}
\sum_{n=1}^{\infty} \left|\frac{\chi(n)}{n^s}\right|
\leq
\sum_{n=1}^{\infty} \frac{1}{n^\sigma}
<\infty.
\end{align*}
Thus $L(s,\chi)$ is defined by an absolutely convergent Dirichlet series.
Now fix a finite set $P$ of primes and define the finite Euler product
\begin{align*}
E_P(\cdot): \{s \in \mathbb{C} : \operatorname{Re}(s)>1\} &\to \mathbb{C} \\
s &\mapsto \prod_{p \in P} \left(1-\chi(p)p^{-s}\right)^{-1}.
\end{align*}
For each $p \in P$, the quantity $\chi(p)p^{-s}$ has modulus at most $p^{-\sigma}<1$, so the geometric series identity gives
\begin{align*}
\left(1-\chi(p)p^{-s}\right)^{-1}
=
\sum_{k=0}^{\infty}\chi(p)^k p^{-ks}.
\end{align*}
Because $P$ is finite, multiplying these geometric series is justified by absolute convergence. Each choice of exponents $(k_p)_{p \in P}$ contributes the term
\begin{align*}
\prod_{p \in P}\chi(p)^{k_p}p^{-k_p s}
=
\frac{\chi\left(\prod_{p \in P}p^{k_p}\right)}
{\left(\prod_{p \in P}p^{k_p}\right)^s},
\end{align*}
where complete multiplicativity of $\chi$ is used in the numerator. Unique factorisation identifies these products exactly with the positive integers whose prime divisors all lie in $P$. Therefore
\begin{align*}
E_P(s)
=
\sum_{\substack{n \geq 1\\ \text{all prime divisors of } n \text{ lie in } P}}
\frac{\chi(n)}{n^s}.
\end{align*}
As $P$ increases over the finite sets of primes, these partial sums exhaust the absolutely convergent Dirichlet series for $L(s,\chi)$. Hence
\begin{align*}
L(s,\chi)
=
\prod_p \left(1-\chi(p)p^{-s}\right)^{-1}.
\end{align*}
[/guided]
[/step]
[step:Differentiate the logarithm of the Euler product term by term]
For each prime $p$, define
\begin{align*}
g_p(\cdot): \{s \in \mathbb{C} : \operatorname{Re}(s)>1\} &\to \mathbb{C} \\
s &\mapsto -\log\left(1-\chi(p)p^{-s}\right),
\end{align*}
where the logarithm is given by the absolutely convergent [power series](/page/Power%20Series)
\begin{align*}
-\log(1-z)=\sum_{k=1}^{\infty}\frac{z^k}{k}
\qquad (|z|<1).
\end{align*}
Thus
\begin{align*}
g_p(s)
=
\sum_{k=1}^{\infty}\frac{\chi(p)^k}{k}p^{-ks}.
\end{align*}
On every closed half-plane $\operatorname{Re}(s)\geq 1+\varepsilon$, with $\varepsilon>0$, the double series
\begin{align*}
\sum_p\sum_{k=1}^{\infty}\frac{\chi(p)^k}{k}p^{-ks}
\end{align*}
converges uniformly and absolutely, since
\begin{align*}
\sum_p\sum_{k=1}^{\infty}\left|\frac{\chi(p)^k}{k}p^{-ks}\right|
\leq
\sum_p\sum_{k=1}^{\infty}p^{-k(1+\varepsilon)}
\leq
\sum_{m=2}^{\infty}\frac{1}{m^{1+\varepsilon}}
<\infty.
\end{align*}
The differentiated double series also converges uniformly and absolutely on the same closed half-plane, because
\begin{align*}
\sum_p\sum_{k=1}^{\infty}\left|(\log p)\chi(p)^k p^{-ks}\right|
\leq
\sum_p\sum_{k=1}^{\infty}(\log p)p^{-k(1+\varepsilon)}
\leq
\sum_{m=2}^{\infty}\frac{\log m}{m^{1+\varepsilon}}
<\infty.
\end{align*}
Therefore term-by-term differentiation of the locally uniformly convergent holomorphic series is valid (citing a result not yet in the wiki: theorem on termwise differentiation of locally uniformly convergent holomorphic series with locally uniformly convergent derivative series). Since
\begin{align*}
\frac{d}{ds}p^{-ks}
=
-k(\log p)p^{-ks},
\end{align*}
we obtain
\begin{align*}
\frac{L'}{L}(s,\chi)
=
\frac{d}{ds}\log L(s,\chi)
=
-\sum_p\sum_{k=1}^{\infty}(\log p)\chi(p)^k p^{-ks}.
\end{align*}
[guided]
The logarithm of the Euler product is handled one prime at a time. For a prime $p$, define
\begin{align*}
g_p(\cdot): \{s \in \mathbb{C} : \operatorname{Re}(s)>1\} &\to \mathbb{C} \\
s &\mapsto -\log\left(1-\chi(p)p^{-s}\right),
\end{align*}
where the logarithm is defined by the power series
\begin{align*}
-\log(1-z)=\sum_{k=1}^{\infty}\frac{z^k}{k}
\qquad (|z|<1).
\end{align*}
This applies because $|\chi(p)p^{-s}|\leq p^{-\sigma}<1$. Substituting $z=\chi(p)p^{-s}$ gives
\begin{align*}
g_p(s)
=
\sum_{k=1}^{\infty}\frac{\chi(p)^k}{k}p^{-ks}.
\end{align*}
We now justify interchanging differentiation with the infinite sums. Fix $\varepsilon>0$ and restrict to the closed half-plane $\operatorname{Re}(s)\geq 1+\varepsilon$. On this set,
\begin{align*}
\sum_p\sum_{k=1}^{\infty}\left|\frac{\chi(p)^k}{k}p^{-ks}\right|
\leq
\sum_p\sum_{k=1}^{\infty}p^{-k(1+\varepsilon)}.
\end{align*}
The last double sum is bounded by the full sum over all integers $m\geq 2$:
\begin{align*}
\sum_p\sum_{k=1}^{\infty}p^{-k(1+\varepsilon)}
\leq
\sum_{m=2}^{\infty}\frac{1}{m^{1+\varepsilon}}
<\infty.
\end{align*}
This proves uniform absolute convergence of the logarithmic series on the closed half-plane.
The derivative of the term $\frac{\chi(p)^k}{k}p^{-ks}$ is
\begin{align*}
\frac{d}{ds}\left(\frac{\chi(p)^k}{k}p^{-ks}\right)
=
-(\log p)\chi(p)^k p^{-ks},
\end{align*}
because $p^{-ks}=e^{-ks\log p}$. The corresponding derivative series is uniformly absolutely convergent on $\operatorname{Re}(s)\geq 1+\varepsilon$, since
\begin{align*}
\sum_p\sum_{k=1}^{\infty}\left|(\log p)\chi(p)^k p^{-ks}\right|
\leq
\sum_p\sum_{k=1}^{\infty}(\log p)p^{-k(1+\varepsilon)}
\leq
\sum_{m=2}^{\infty}\frac{\log m}{m^{1+\varepsilon}}
<\infty.
\end{align*}
The last series converges for $\varepsilon>0$. Hence the usual holomorphic termwise differentiation theorem applies to the locally uniformly convergent series (citing a result not yet in the wiki: theorem on termwise differentiation of locally uniformly convergent holomorphic series with locally uniformly convergent derivative series).
Therefore
\begin{align*}
\frac{d}{ds}\log L(s,\chi)
=
-\sum_p\sum_{k=1}^{\infty}(\log p)\chi(p)^k p^{-ks}.
\end{align*}
Since $\frac{d}{ds}\log L(s,\chi)=L'(s,\chi)/L(s,\chi)$ wherever the Euler product is nonzero, and the Euler product has no zero because none of its local factors vanishes, this gives
\begin{align*}
\frac{L'}{L}(s,\chi)
=
-\sum_p\sum_{k=1}^{\infty}(\log p)\chi(p)^k p^{-ks}.
\end{align*}
[/guided]
[/step]
[step:Reindex the prime-power series by the von Mangoldt function]
Multiplying by $-1$ gives
\begin{align*}
-\frac{L'}{L}(s,\chi)
=
\sum_p\sum_{k=1}^{\infty}(\log p)\chi(p)^k p^{-ks}.
\end{align*}
If $n=p^k$ for a prime $p$ and $k \in \mathbb{N}$, then complete multiplicativity of $\chi$ gives $\chi(n)=\chi(p)^k$, and by definition $\Lambda(n)=\log p$. If $n$ is not a prime power, then $\Lambda(n)=0$. Hence the coefficient of $n^{-s}$ in the right-hand side is exactly $\Lambda(n)\chi(n)$ for every $n \in \mathbb{N}$. Therefore
\begin{align*}
-\frac{L'}{L}(s,\chi)
=
\sum_{n=1}^{\infty}\frac{\Lambda(n)\chi(n)}{n^s}.
\end{align*}
The absolute convergence of this series follows from the preceding absolute convergence of the prime-power series, so the reindexing is legitimate. This proves the formula for every $s$ with $\operatorname{Re}(s)>1$.
[/step]
