Value of cyclotomic polynomial evaluated at 1 Let $\Phi_n(x)$ be the usual cyclotomic polynomial (minimal polynomial over the rationals for a primitive nth root of unity).
There are many well-known properties, such as $x^n-1 = \Pi_{d|n}\Phi_d(x)$.
The following fact appears to follow pretty easily:
Fact:
$\Phi_n(1)=p$ if $n$ is a prime power $p^k$.
$\Phi_n(1)=1$ if $n$ is divisible by more than one prime.
My question is, is there a reference for this fact? Or is it simple enough to just call it "folklore" or to just say it "follows easily from properties of cyclotomic polynomials".
 A: Another proof follows directly from the formula $X^{n} - 1 = \prod_{d \mid n} \Phi_d(x)$, since we can deduce from it that 
\begin{equation}
X^{n-1} + \cdots + X + 1 = \prod_{d \mid n, d>1} \Phi_d(x).
\end{equation} 
Thus, if $n = p^{k}$, we have 
$$
X^{p^{k}-1} + \cdots + X + 1 = \Phi_{p}(x) \cdots \Phi_{p^{k-1}}(x) \Phi_{p^{k}}(x).
$$
After evaluating in 1 we obtain $p^{k} = \Phi_{p}(1) \cdots \Phi_{p^{k-1}}(1) \Phi_{p^{k}}(1)$ and induction on $k$ gives $\Phi_{p^{k}}(1) = p$ for all $k$.
If $n = p_{1}^{\alpha_{1}} \cdots  p_{r}^{\alpha_{r}}$, where $\alpha_{i}$'s are positive integers and $r \geq 2$, then 
$$
n = \Phi_{n}(1) \prod_{d \mid n, d\neq 1,n} \Phi_d(1).
$$
If we assume the statement true for all positive integers $<n$ then the product in the left member of the equation equals $n$, since 
$$
\prod_{i=1}^{r}\Phi_{p_{i}}(1) \cdots \Phi_{p_{i}^{\alpha_{i}}}(1) = p_{1}^{\alpha_{1}} \cdots p_{r}^{\alpha_{r}} = n
$$
and the rest of the factors are 1. Thus, $\Phi_{n}(1) = 1$ also.
A: Möbius Inversion:
As outlined in Qiaochu's comment, Möbius inversion will solve this problem.  Since I am more comfortable with sums then products, lets just take logs.  We have $$\log n=\sum_{d|n\ d\neq 1}\log\Phi_{d}(1).$$  Then for $d\neq1$, $$\log\Phi_{d}(1)=\sum_{d|n}\mu\left(\frac{n}{d}\right)\log d=\Lambda(n)$$ where $\Lambda(n)$ is the Von Mangoldt Lambda Function.  Since $\Lambda(p^k)=\log p$, and $\Lambda(n)=0$ for $n$ composite, the result then follows upon exponentiating. 
Other: 
This relation follows from some other identities.  For an integer $n$ and a prime $p$ we have that $$\Phi_{np}(x)=\frac{\Phi_{n}\left(x^{p}\right)}{\Phi_{n}(x)}\ \text{when }\gcd(n,p)=1$$
$$\Phi_{np}(x)=\Phi_{n}\left(x^{p}\right)\ \text{when }\gcd(n,p)=p.$$
We know that $\Phi_p(1)=p$, and from the above it follows that $\Phi_{p^\alpha}(1)=p$ and $\Phi_{pq}(1)=1$. 
Hope that helps,
