Product of all reduced residues in relation with the function $\mu$: $\prod_{1\leq a\leq n,(a,n)=1}a=n^{\varphi(n)}\prod_{d|n}(d!/d^d)^{\mu(n/d)}$

Question

We know that for any arithmetic function $f$ the Mobius inversion formula gives its inversion. Hence $$ F(n)=\prod_{d|n}f(d)\implies f(n)=\prod_{d|n} F(n/d)^{\mu(d)}.$$ The above statement can be proven by taking logarithm on both sides. Using this result, I think showing the result as stated in the title $$ \prod_{1\leq a\leq n,(a,n)=1}a=n^{\varphi(n)}\prod_{d|n}\left(\frac{d!}{d^d}\right)^{\mu(n/d)} $$ is equivalent to showing $$ \frac{n!}{n^n}=\prod_{d|n}\prod_{1\leq a\leq d, (a,d)=1}\frac{a}{d}. $$ However, I cannot proceed from here efficiently. Is this the right track to the solution and is there any hints/key observations that I am missing?