The following question is on page 48 apostol Introduction to analytic number theory( 20)
Let $P(n)$ be the product of the positive integers which are less than equal to $n$ and relatively prime to $n$.
Then prove that $ P(n) = n^{\phi(n) } \prod_{d|n} {(\frac{ d! }{d^d})}^{\mu(n/d) } $ .
I tried by manipulating RHS and noting that function on RHS is multiplicative but I have been not successful.
So, can you please help me.
Simple rearrangment followed by taking logarithm of both sides yields $$S(n) = \sum_{m \leq n, (m,n) = 1}\log\bigg(\dfrac{m}{n} \bigg) = \sum_{d|n}\mu(n/d)T(d)$$ Where $T(n) = \log (n!/n^n)$. Now by applying the mobius inversion formula $S = \mu * T$ if and only if $T = S*u$. That is $$\log(n!/n) = \sum_{d|n} \sum_{m\leq d, (m,d) = 1} \log(m/d)$$
Hence it suffices to show that $$\dfrac{n!}{n^n} = \prod_{d|n}\prod_{m\leq d, (m,d) = 1}\dfrac{m}{d} $$.
This simpler to work with. One way to show the equation is true uses the following lemma.
The set $$S = \{ m\cdot n/d | d \text{ divides } n,\ 1 \leq m \leq d,\ (m,d) = 1 \}$$ runs through every number modulo $n$ exactly once.\
You can show that this lemma is true easily(Hint: Show that $S \subseteq \{1, \dots , n\}$ and $\{1, \dots , n\} \subseteq S$.)
Returning back to that last product formula, you can see that $$\prod_{d|n}\prod_{m\leq d, (m,d) = 1}\dfrac{m}{d} = \prod_{d|n}\prod_{m\leq d, (m,d) = 1}\dfrac{mn/d}{n} = \dfrac{n!}{n^n} $$