Proving $P(n) =n^{\phi(n)} \prod\limits_{d \mid n} \left(\frac{d!}{d^d} \right)^{\mu(n/d)}$
Solution 1:
Success finally!
Let,
$$f( n) = \sum_{(k,n)=1;1\leq k\leq n} \log\Bigl(\frac{k}{n}\Bigr)$$ therefore we have
$$\sum_{d|n}f(d) =\log\Bigl(\frac{1}{n}\Bigr)+...+\log\Bigl(\frac{n}{n}\Bigr)=\log\left(\frac{n!}{n^n}\right)$$
Thus by Moebius Inversion Formula:
$$f(n) = \sum_{d|n}\log\left(\frac{d!}{d^d}\right)\cdot \mu\left(\frac{n}{d}\right) = \log\left(\prod_{d|n}\left(\frac{d!}{d^d}\right)^{\mu\left(\frac{n}{d}\right) }\right)$$
$$f(n) = \sum_{(k,n)=1;1\leq k\leq n} {\log(k)} -\phi(n)\cdot \log( n) = \log(P(n))-\log(n^{\phi(n)})$$