Value of cyclotomic polynomial evaluated at 1

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.

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,