Prov ∑d|n mu(d)/d= (1-1/p_1)(1-1/p_2)...(i-p_r)?

Note that since μ and identity functions are multiplicative, we have μ(d)/d is multiplicative.
Hence, Σ_(d|n) μ(d)/d is also multiplicative.

So, it suffices to compute this sum when n = p^k for some prime p and posiitive integer k.

In this case,
Σ_(d|p^k) μ(d)/d
= μ(1)/1 + μ(p)/p + μ(p^2)/p^2 + ... + μ(p^k)/p^k
= 1/1 + (-1)/p + 0/p^2 + ... + 0/p^k
= 1 - 1/p.
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So, if we write n = (p1)^(a1) ... (pr)^(ar) as the prime factorization of n
(with p1, ..., pr distinct primes), we conclude that
Σ_(d|n) μ(d)/d = Π(k = 1 to r) (1 - 1/p_r).

I hope this helps!