Dirichlet Convolution of Mobius function and distinct prime factor counter function. Let us define an Arithmetical function $\nu(1)=0$. For $n > 1$, let $\nu(n)$ be the number of distinct prime factors of $n$.
I need to prove $\mu * \nu (n)$ is always 0 or 1.
According to my computation, if $n$ is prime, it is 1. If $n$ is composite, it is 0.
I tried using induction, so that assuming for a composite number $m$, $\mu*\nu(m) = 0$, then for any prime $q$, $\mu*\nu(qm) = 0$.

$$F(a,s) = \prod_p (1+a \sum_{k=1}^\infty p^{-sk}) = \sum_{n=1}^\infty n^{-s} a^{\nu(n)}$$
$$\frac{\partial F(a,s)}{\partial a}|_{a=1} = \sum_{n=1}^\infty n^{-s} \nu(n)$$
$$G(a,s) = \frac{F(a,s)}{\zeta(s)} = \prod_p (1-p^{-s}) (1+a \sum_{a=1}^\infty p^{-sk}) = \prod_p (1+(a-1)p ^{-s})
$$
$$\frac{\partial G(a,s)}{\partial a}|_{a=1} = G(1,s)\sum_p \frac{\partial (1+(a-1)p ^{-s})}{\partial a}|_{a=1} = G(1,s) \sum_p p^{-s} = \sum_p p^{-s}$$
thus $\nu \ast \mu (n) = \delta_\pi(n)$ where $\delta_\pi(n)$ is the prime indicating function
