If $\zeta$ is a primitive $n$-th root of unity, prove that: $$d(1, \zeta,...,\zeta^{\varphi(n)-1})=(-1)^{\varphi(n)/2}n^{\varphi(n)}\prod_{p\mid n} p^{-\frac{\varphi(n)}{p-1}}$$
Let $n=\prod_{i=1}^{m}p_i^{e_i}$. After looking it up in some books, I was able to understand why this is true for $m=1$. However, they all ignored the general case $m>1$ or simply stated that it could be done my induction on $m$, but I really can't see how it could be done.
The only interesting thing I could find out was that for $n,m$ with $\gcd(n, m)=1$, we get, on the right hand side of the equation:
$$(-1)^{\varphi(nm)/2}nm^{\varphi(nm)}\prod_{p\mid nm} p^{-\frac{\varphi(nm)}{p-1}}=$$ $$\left((-1)^{\varphi(n)/2}n^{\varphi(n)}\prod_{p\mid n} p^{-\frac{\varphi(n)}{p-1}}\right)^{\varphi(m)}\left((-1)^{\varphi(m)/2}n^{\varphi(m)}\prod_{p\mid m} p^{-\frac{\varphi(m)}{p-1}}\right)^{\varphi(n)}$$
That makes me think I'm getting somewhere, but I'm stuck with the problem of showing that $d(1, \zeta,...,\zeta^{\varphi(nm)-1})=[d(1, \zeta,...,\zeta^{\varphi(n)-1})]^{\varphi(m)}[d(1, \zeta,...,\zeta^{\varphi(m)-1})]^{\varphi(n)}$, which doesn't seem trivial at all. Any ideas? Thanks!