$P(n) = n^{ \phi(n)} \prod_{d|n}{(d!/d^d)}^{\mu(n/d)} $ Let $P(n)$ be the product of the positive integers which are $\leq n$ and relatively
prime to $n$. Prove that
  $$P(n) =  n^{\phi(n)} \prod_{d|n}{(d!/d^d)}^{\mu(n/d)}. $$  
 A: We will use the following well-known identity:
Lemma 1
$$
\phi(n) = \sum_{d|n} d\mu(\frac nd).
$$ 
From this, the identity in the question follows: 
Proof
By inclusion-exclusion principle, 
$$
\begin{align}
\log P(n)&= \sum_{d|n} \mu(d) \left( \log d + \log 2d + \cdots + \log \left(\frac  nd  d \right)\right)\\
&=\sum_{d|n} \mu(d) \left(\log (\frac nd)! + \frac nd \log d\right)\\
&=\sum_{d|n} \mu(\frac nd)\left( \log d! +  d \log \frac nd \right)\\
&=\phi(n)\log n + \sum_{d|n} \left( \mu(\frac nd )\log d! - d\mu(\frac nd)\log d.  \right)
\end{align}
$$
Therefore, 
$$
P(n) = n^{\phi(n)} \prod_{d|n} \left(\frac {d!}{d^d}\right)^{\mu(\frac nd)}.
$$
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.
