$P(n) = n^{ \phi(n)} \prod_{d|n}{(d!/d^d)}^{\mu(n/d)} $

Question

Let $P(n)$ be the product of the positive integers which are $\leq n$ and relatively prime to $n$. Prove that $$P(n) = n^{\phi(n)} \prod_{d|n}{(d!/d^d)}^{\mu(n/d)}. $$

Answer 1

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, $$ P(n) = n^{\phi(n)} \prod_{d|n} \left(\frac {d!}{d^d}\right)^{\mu(\frac nd)}. $$

Answer 2

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.