Inverse of Dirichlet series equality I stumbled across a formula in here
and tried to prove it for myself: $$\frac{1}{L(s,\chi)}=\sum\limits_{n=1}^{\infty}\frac{\mu(n)\chi(n)}{n^s}$$ However I got stuck. In my attempt I tried to show that: $$\frac{1}{L(s,\chi)}=\prod\limits_{p}(1-\chi(p)\ p^{-s})$$ While $$\sum\limits_{n=1}^{\infty}\frac{\mu(n)\chi(n)}{n^s}=\prod\limits_{p}(1-\chi(p)\mu(p)\ p^{-s})^{-1}= \prod\limits_{p}(1+\chi(p)\ p^{-s})^{-1}$$Then I tried to show that the terms in the product are equal but that doesn't seem true. What am I missing here? 
 A: Recall  that the  product of  two Dirichlet  series translates  into a
Dirichlet convolution of their coefficients.
In the present case we seek to obtain that
$$1 = L(s,\chi) \times \sum_{n\ge 1} \frac{\mu(n)\chi(n)}{n^s}$$
so we must show that
$$\sum_{d|n} \mu(d)\chi(d)\chi(n/d)
= \begin{cases}
1 & \text{if} \quad n=1 \\
0 & \text{otherwise}.
\end{cases}.$$
The part where $n=1$ follows by inspection.

Note  that we may  assume in  this convolution  that $\gcd(n,k)  = 1,$
where $k$  is the  period of the  character, because otherwise at least one of
$\chi(d)$ or  $\chi(n/d)$ would always  end up not being  coprime with
$k$, rendering the sum zero.

Now let $p^v$  represent the prime factors $p$ of  $n$ with their exponent
$v.$ By virtue of $\chi$ being completely multiplicative we then have
$$\chi(d)\chi(n/d) = \prod_{p|n} \chi(p)^v$$
and the convolution becomes
$$\sum_{d|n} \mu(d) \times \prod_{p|n} \chi(p)^v.$$
The first factor is zero when $n>1$, which concludes the argument.
A: The factorization that you've written down,
$$\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=\prod \left(1-\color{Red}{\mu(p)}p^{-s}\right)^{\color{Red}{-1}},$$
is incorrect. The correct factorization is
$$\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=\prod \left(1-p^{-s}\right)$$
Notice how the negative signs within each $p$-factor are enough to get the $\mu$ factors in the resulting series expansion, and so there is no $\mu(p)$ to write down. There is also no $-1$ power. Expanded:
$$\small 1-\left(\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{5^s}+\cdots\right)+\left(\frac{1}{6^s}+\frac{1}{10^s}+\frac{1}{14^s}+\cdots\right)-\cdots=\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{5^s}\right)\cdots $$
This is no different than $1-(x+y+z)+(xy+yz+xz)-(xyz)=(1-x)(1-y)(1-z)$ for instance, only with infinitely many variables and powers of primes in place of variables.
A: It's true that if $f(n)$ is a completely multiplicative function, then
$$
\sum_{n=1}^\infty \frac{f(n)}{n^{s}} = \prod_p (1-f(p)p^{-s})^{-1}.
$$
For a function that is merely multiplicative, the general formula is
$$
\sum_{n=1}^\infty \frac{f(n)}{n^{s}} = \prod_p (1+f(p)p^{-s}+f(p^2)p^{-2s}+\cdots)
$$
(from which you recover the first formula when $f(p^k)=f(p)^k$ for all $k$).
You tried applying the first formula to the multiplicative function $f(n)=\mu(n)\chi(n)$, but that function is not completely multiplicative. The second formula should make you happier.
Of course, this exercise should also be paired with understanding deeply why the second formula is true!
