Cyclotomic Polynomial Evaluated at 1

Question

I noticed that when trying to evaluate the $n$th cyclotomic polynomial at $1$, $\Phi_n(1)$, we run into an issue while using the explicit formula $\Phi_n(x)=\prod_{d|n}(x^d-1)^{\mu(n/d)}$. In particular, $$\Phi_n(1)=\prod_{d|n}(1^d-1)^{\mu(n/d)}=\prod_{d|n}0=0?$$ This is certainly not true, since $\Phi_n(1)$ is $1$ for all integers $n>1$ such that $n$ is not a power of a prime and $\Phi_n(1)=p$ if $n=p^k$ (Proved as a lemma here). Can someone explain what's going on here? Any help is appreciated.

Answer 1

The answer lies in $\mu(n/d)$. In particular $\mu(p) = -1$ for any prime p. Since n/d is certainly a prime number for some choice of d, we have that $\mu(n/d) = -1$ for some choice of d|n. Then we have a rational function whose denominator evaluates to 0 at 1, so $x = 1$ isn't in the domain of this function.

Hope that helps.