Prove that $f_n(z)= \prod_{k|n} (z^k-1)^{\mu(n/k)}$ I have the following problem and I would really like some hint or suggestion on where to start working since I have no idea. 
The problem is as follows: Let $f_n(z)$ be the function that has as its zeros all numbers $\eta$ for which $\eta^n=1$ but $\eta^k \neq 1$ for $1 \leq k <n$. Prove that $f_n(z)= \prod_{k|n} (z^k-1)^{\mu(n/k)}$.
Thanks for the help and suggestions!
 A: First of all I suggested you to pay attention to the hypothesis, 
because there are infinitely many functions $f_{n}(z)$ with your properties;
I think that you ask about a polynomial one, i.e about $f_{n}(z)=\phi_{n}(z)=\displaystyle \prod_{\zeta}(z-\zeta)$,the $n-th$ cyclotomic polynomials, with $\zeta$ that goes over all the primitive $n-th$ root of unity, i.e. the number that satisfies $\zeta^{n}=1,\zeta^{k}\neq 1,\forall 1\leq k<n$.
In this case is quite easy to find that $z^{n}-1=\displaystyle \prod_{k|n}\phi_{k}(z)$ because $z^{n}-1=\displaystyle \prod_{k|n}\prod_{\zeta^{k}=1}(z-\zeta)$, since $z^{n}-1$ have as root all the $n-th$ root of unity and a generic $n-th$ root of unity $\zeta$ satisfy $\zeta^{k}=1$ for some $k|n$ (essentially for Lagrange theorem applied to the group of all the $n-th$ root of unity).
Now you can procede by apply the Mobius inversion formula (for the product) to this equality and to find that $\phi_{n}(z)=\displaystyle \prod_{k|n}(z^{k}-1)^{\mu(\frac{n}{k})}$.
