Proving $P(n) =n^{\phi(n)} \prod\limits_{d \mid n} \left(\frac{d!}{d^d} \right)^{\mu(n/d)}$ Actually, I posted this long ago in MO but did not get a reply as it was unfit. 
Now this is an exercise in some textbook (I think Apostol), and I would be happy to receive some answers.
Let $P(n)$ be the product of positive integers which are $\leq n$ and relatively prime to $n$.  Prove that $$ \displaystyle P(n) = n^{\phi(n)} \prod\limits_{d \mid n} \left(\frac{d!}{d^d} \right)^{\mu(n/d)}.$$
 A: 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: 
$$f(n) = \sum_{d|n}\log\left(\frac{d!}{d^d}\right)\cdot \mu\left(\frac{n}{d}\right) = \log\left(\prod_{d|n}\left(\frac{d!}{d^d}\right)^{\mu\left(\frac{n}{d}\right) }\right)$$
$$f(n) = \sum_{(k,n)=1;1\leq k\leq n} {\log(k)} -\phi(n)\cdot \log( n) = \log(P(n))-\log(n^{\phi(n)})$$
A: Start by classifying $[n]$ according to GCD:
$$n! = \prod_{d|n} \prod_{(q,n)=d} q$$
where $q$ ranges from $1$ to $n.$ This is
$$n! = \prod_{d|n} \prod_{(r,n/d)=1} (dr)
= \prod_{d|n} d^{\varphi(n/d)} \prod_{(r,n/d)=1} r
\\ = \prod_{d|n} d^{\varphi(n/d)} P(n/d). $$
This becomes
$$n! = \prod_{d|n} (n/d)^{\varphi(d)} P(d)
= \prod_{d|n} n^{\varphi(d)} \prod_{d|n} d^{-\varphi(d)} P(d)
\\ = n^n \prod_{d|n} d^{-\varphi(d)} P(d).$$
so that we find 
$$\prod_{d|n} d^{-\varphi(d)} P(d) = \frac{n!}{n^n}.$$
By Mobius inversion we thus have
$$n^{\Large -\varphi(n)} P(n) = 
\prod_{d|n} \left(\frac{d!}{d^d}\right)^{\Large \mu(n/d)}.$$
This finally yields
$$\bbox[5px,border:2px solid #00A000]{
P(n) = n^{\Large \varphi(n)} 
\prod_{d|n} \left(\frac{d!}{d^d}\right)^{\Large \mu(n/d)}.}$$
as claimed.
