Dirichlet Series with Dirichlet exponential 
Let $F(s) = \sum_{n=1}^{\infty}f(n)n^{-s}$ where $f(n)$ is completely multiplicative and the series converge for $\text{Re}> \sigma_0 $. Show that for $\text{Re}> \sigma_0 $ $$\frac{F^{'}(s)}{F(s)} = \sum_{n=1}^{\infty} \frac{f(n) \Lambda(n)}{n^{s}}.$$ 

Is there a proof of this without using Dirichlet exponential? 
The only proof I've known about this is using Dirichlet exponential
(See https://www.springer.com/gp/book/9780387901633).
 A: 
If $f(n)$ is completely multiplicative and $\sum_{n=1}^\infty f(n) n^{-s}$ converges absolutely for some $s_0$ then for $\Re(s) > \Re(s_0)$ 

$$-F'(s)= \sum_{n=1}^\infty f(n) n^{-s}\log n ,\frac1{F(s)}= \sum_{n=1}^\infty f(n) \mu(n) n^{-s}, \frac{-F'(s)}{F(s)}= \sum_{n=1}^\infty f(n) \Lambda(n) n^{-s}$$
The absolute convergence is needed, otherwise you would obtain that $\sum_{n=1}^\infty \chi(n) n^{-s}$ has no zeros for $\Re(s) > 0$ for any non-trivial Dirichlet character.

We define $\Lambda(n) = \log p$ if $n=p^k$ and $0$ otherwise.

The absolute convergence of the above series is obvious thus it suffices to show the equalities are true as formal Dirichlet series.
$$(\sum_{m=1}^\infty f(m) m^{-s})(\sum_{d=1}^\infty f(d) \Lambda(d) d^{-s}) = \sum_{n=1}^\infty  n^{-s} \sum_{n=md} f(d) \Lambda(d)f(m)$$ $$=\sum_{n=1}^\infty  n^{-s} f(n) \sum_{p^k | n} \log p=\sum_{n=1}^\infty  n^{-s} f(n) \log n=-F'(s) $$
Whence $$\frac{-F'(s)}{F(s)}=\sum_{n=1}^\infty f(n) \Lambda(n) n^{-s}$$
Defining $\mu(n)$ as the Dirichlet inverse of $1_{n=1}$ the proof for $1/F(s)$ is similar.
