For an integer $n \geq 1$, verify the formula: $\sum\limits_{d|n} \mu (d) \lambda(d)= 2^{\omega (n)}$ For an integer $n \geq 1$, verify the formula:
$\sum\limits_{d|n} \mu (d) \lambda(d)= 2^{\omega (n)}$ 
I know that this problem somehow uses the Mobius Inversion Formula but I am VERY confused how to use it.
 A: Let  $$F(n) = \sum\limits_{d\mid n} \mu(d)f(d)$$
Observe that $F(p^k)= 1-f(p)$ because $\mu(p^i) = 0$ for all $i\gt 1$.
Also since $F(n)$ is multiplicative, for $n=p_1^{k_2}p_2^{k_2}\cdots p_r^{k_r}$ we have$$F(n) = (1-f(p_1))(1-f(p_2))\cdots (1-f(p_r))\tag{1}$$
Plugging in $f = \lambda $ we get the desired result
$$\begin{align} \sum\limits_{d\mid n} \mu(d)\lambda(d)&=(1-\lambda(p_1))(1-\lambda(p_2))\cdots (1-\lambda(p_r))\\~\\&=(1-(-1))(1-(-1))\cdots (1-(-1))\\~\\&=(2)(2)\cdots(2)\\~\\&=2^{\omega(n)}\end{align}$$
A: Suppose we seek to show that
$$\sum_{d|n} \mu(d) \lambda(d) = 2^{\omega(n)}.$$
This can be done using Dirichlet series and Euler products.
We have for the RHS and $$2^{\omega(n)}$$ the Euler product
$$\prod_p 
\left(1 + \frac{2}{p^s} + \frac{2}{p^{2s}} + \frac{2}{p^{3s}}
+\cdots\right).$$
which is
$$\prod_p \left(-1 + 2\frac{1}{1-1/p^s}\right)
= \prod_p \frac{-1+1/p^s+2}{1-1/p^s}
\\ = \prod_p \frac{1+1/p^s}{1-1/p^s}
= \prod_p \frac{1-1/p^{2s}}{(1-1/p^s)^2}
= \frac{\zeta(s)^2}{\zeta(2s)}.$$
On the other hand, we get for $$\mu(n)\lambda(n)$$ the Euler product
$$\prod_p 
\left(1 + \frac{(-1)\times (-1)}{p^s} \right).$$
which is
$$\prod_p \left(1 + \frac{1}{p^s} \right)
= \prod_p \frac{1-1/p^{2s}}{1-1/p^{s}}
= \frac{\zeta(s)}{\zeta(2s)}$$
and  hence  for  the  LHS   (summing  over  the  divisors  of  $n$  is
multiplication by $\zeta(s)$)
$$\zeta(s)\times \frac{\zeta(s)}{\zeta(2s)}
= \frac{\zeta(s)^2}{\zeta(2s)}$$
and   we   have  equality   of   the   Dirichlet   series  and   their
coefficients.   Since  $\omega(n)   <\log_2  n$   and   since  $\sum_p
\frac{1}{p^2}$ converges  the half-plane of convergence  for these two
is $\Re(s)> 2.$
Remark. If  we want to be  rigorous about it, we  have equality of
the coefficients  of the Dirichlet  series in the intersection  of the
two  half-planes  of  convergence  because for  the  Dirichlet  series
$\Lambda(s) = \sum_{n\ge 1} \frac{\lambda_n}{n^s}$ we have
$$\lambda_n = \frac{1}{2\pi i} 
\int_{c-i\infty}^{c+i\infty} \Lambda(s) n^{s-1} ds.$$
Addendum. An elementary argument is to notice that the
LHS is the sum
$$\sum_{Q\subseteq P} (-1)^{|Q|} \times (-1)^{|Q|}
= \sum_{Q\subseteq P} 1 = 2^{\omega(n)}$$
which is the RHS. Here $P$ is the set of primes dividing $n$
so that $\omega(n) = |P|.$
This  is  because  the  Moebius  function  is zero  when  $n$  is  not
squarefree and we may sum over  all genuine subsets of $P$ (as opposed
to multisets). These subsets correspond to squarefree divisors.
The first $(-1)^{|Q|}$ is the contribution from the Moebius function
and the second one from the Liouville function.
