# Proving $P( n) =n^{\phi(n)} \prod\limits_{d \mid n} \biggl(\frac{d!}{d^d} \biggr)^{\mu(n/d)}$

Actually, i had posted this long ago in MO, and i didn't get a reply to this question as it was unfit.

Now, this is an exercise, in some textbook ( i think Apostol) and i would be happy if i can receive some answers.

Let $P(n)$ be the product of positive integers which are $\leq n$ and relatively prime $n$. Prove that $$\displaystyle P( n) =n^{\phi(n)} \prod\limits_{d \mid n} \biggl(\frac{d!}{d^d} \biggr)^{\mu(n/d)}$$

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This is screaming for you to use the Moebius inversion formula. – Mariano Suárez-Alvarez Oct 22 '10 at 18:52
@Mariano: Thanks! But i actually couldn't figure out as to what my $f(n)$ and $g(n)$ should be! – anonymous Oct 22 '10 at 18:54
Looking at the equation you want to prove, it is clear that there is exactly one choice. Maybe if you wrote what you tried, we could help you. Otherwise, I'd be just ruining your problem for you. – Mariano Suárez-Alvarez Oct 22 '10 at 18:57
@Mariano: I think i have got it now.! – anonymous Oct 30 '10 at 20:34

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:

$$f(n) = \sum_{d|n}\log\left(\frac{d!}{d^d}\right)\cdot \mu\left(\frac{n}{d}\right) = \log\left(\prod_{d|n}\left(\frac{d!}{d^d}\right)^{\mu\left(\frac{n}{d}\right) }\right)$$

$$f(n) = \sum_{(k,n)=1;1\leq k\leq n} {\log(k)} -\phi(n)\cdot \log( n) = \log(P(n))-\log(n^{\phi(n)})$$

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