In text book Analytic number theory by Apostol on page $47$, Exercise $5$ we have the following:
Define $v(1)=0$, and for $n>1$ let $v(n)$ be the number of distinct prime factors of $n$. Let $f=\mu*v$ and prove that $f(n)$ is either $0$ or $1$.
The symbol $*$ is taken as the Dirichlet product.
I tried to divide cases into whether $n$ is square free or not. But I can't find anything. Please help me to solve it.
On
This can also be done using Dirichlet series. Start with
$$L_1(s) = \sum_{n\ge 1}\frac{\mu(n)}{n^s} = \prod_p\left(1-\frac{1}{p^s}\right) = \frac{1}{\zeta(s)}.$$
Furthermore introduce
$$L_2(s) = \sum_{n\ge 1}\frac{\omega(n)}{n^s} = \left. \left(\prod_p\left(1+\frac{a}{p^s}+\frac{a}{p^{2s}}+\cdots\right) \right)'\right|_{a=1}.$$
Here the derivative is with respect to $a.$ Continuing,
$$\left. \left(\prod_p\left(1+a\frac{1/p^s}{1-1/p^s}\right) \right)'\right|_{a=1} \\ = \left. \left(\prod_p\left(1+a\frac{1/p^s}{1-1/p^s}\right) \sum_p \frac{\frac{1/p^s}{1-1/p^s}} {1+a\frac{1/p^s}{1-1/p^s}} \right)\right|_{a=1} \\ = \zeta(s) \sum_p \frac{1/p^s}{1-1/p^s+1/p^s} = \zeta(s) \sum_p \frac{1}{p^s}.$$
In fact this last equation could have been obtained by inspection. It follows that
$$L(s) = \sum_{n\ge 1} \frac{(\mu\star\omega)(n)}{n^s} \\ = L_1(s) L_2(s) = \sum_p \frac{1}{p^s} = \sum_{n\ge 1} \frac{1}{n^s} [[n\;\text{is prime}]]$$
as claimed.
Let us call $\omega\left(n\right) $ the function that count the number of distinct prime factors of $n$. We have that $$\sum_{d\mid n}\delta_{p}\left(d\right)=\omega\left(n\right) $$ where $$\delta_{p}\left(d\right)=\begin{cases} 1, & d=p,\, p\textrm{ is a prime number}\\ 0, & \textrm{otherwise} \end{cases} $$ and we have done since, by Möbius inversion formula, we get $$f\left(n\right)=\sum_{d\mid n}\mu\left(d\right)\omega\left(\frac{n}{d}\right)=\delta_{p}\left(n\right).$$