Möbius Inversion with Liouville Function

This article concerns Möbius Inversion with the Liouville Function.  For ` n=p_1^{a_1}p_2^{a_2}...p_k^{a_k} `, the number of distinct prime factors of n is

Question
(i)  Prove that

(ii) Hence show that

Solution:-
(i) Note that ` \mu(d)=0 ` for divisors d containing any prime power of index 2 or higher.  We only need to consider divisors of the form ` d=p_{i_1}p_{i_2}...p_{i_h} `, where ` h = 0` is interpreted as ` d=1 `.
    ` \sum_{d|n}\mu(d)\lambda(d) `
= ` \sum_{h=0}^{k}\sum_{p_{i_1},p_{i_2},...,p_{i_h}} \mu(p_{i_1}p_{i_2}...p_{i_h})\lambda(p_{i_1}p_{i_2}...p_{i_h}) `
= ` \sum_{h=0}^{k}\sum_{p_{i_1},p_{i_2},...,p_{i_h}} (-1)^h(-1)^{1+1+...+1} `   (where 1+1+...+1 has h copies of 1)
= ` \sum_{h=0}^{k}\sum_{p_{i_1},p_{i_2},...,p_{i_h}} (-1)^{2h} ` = ` \sum_{h=0}^{k}\sum_{p_{i_1},p_{i_2},...,p_{i_h}} 1 ` = ` \sum_{h=0}^{k}((k),(h)) `
= ` 2^k `  = ` 2^{\omega(n)} `
(ii)  Recall that ` \lambda(\frac{n}{d})=\frac{\lambda(n)}{\lambda(d)}=\lambda(n)\lambda(d) `
     From the result in part (i) we multiply both sides by ` \lambda(n) ` to get
                  ` \sum_{d|n}\mu(d)\lambda(n)\lambda(d)= \lambda(n)2^{\omega(n)} `
                  ` \sum_{d|n}\mu(d)\lambda(\frac{n}{d})= \lambda(n)2^{\omega(n)} `
    Treating this as the "Möbius Inversed" formula, the "original" formula is
                  ` \sum_{d|n}\lambda(d)2^{\omega(d)}= \lambda(n) `
Hence
     ` \sum_{d|n}\lambda(\frac{n}{d})2^{\omega(d)} ` =   ` \sum_{d|n}\lambda(n)\lambda(d)2^{\omega(d)} ` =   ` \lambda(n)\cdot\sum_{d|n}\lambda(d)2^{\omega(d)} `
=   ` \lambda(n)\cdot\lambda(n) ` =   ` \lambda(n)^2 ` =   ` 1 `

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