Hint: For $n>1$, $\Phi_n(x)=\frac{\sum_{k=0}^{n-1}x^k}{\prod_{1 < d < n,~d|n}\Phi_d(x)}$ is the ratio of $\frac{x^n-1}{x-1}$ to the product of $\Phi_d(x)$ over all proper divisors $d$ of $n$. The numerator is just $n$ when $x=1$ (which also follows from L'Hopital's rule for the limit of the ratio as $x\to1$), and the denominator gives you an inductive statement about $a_n=\left\{\matrix{1&n=1~\\\Phi_n(1)&n>1,}\right.$ namely $ a_1=1,\quad a_n=\frac{n}{\prod_{1 where, again, the product is over all proper divisors $d$ of $n$. For example the result that $a_n=p$ for $n=p^k$ a positive power of a prime follows by an easy induction.
For the general case, write $a(n)=\Phi_n(1)+\epsilon(n)$, as a function, where $\epsilon(n)=\delta_{1n}$ is $1$ for $n=1$ and $0$ for $n\ne1$. Then $a$ is an arithmetic function, in fact a unit, and $\epsilon$ is the multiplicative identity, in the Dirichlet ring of such functions, and we have the identity $n=\prod_{0 (i.e., $\log(n)=\log(a(n))*1$ using Dirichlet convolution and natural logarithms). This has solution given by the Möbius inversion formula, which states that $f=g*1\iff g=f*\mu$ for arithmetic functions $f,g$: $ \eqalign{ \log(a(n))& =\sum_{01}\right. }$ so that $ \Phi_n(1)=\left\{\matrix{ 0&\text{if }n=1~~\\ p&\text{if }n=p^k\\ 1&\omega(n)>1}\right. $
EDIT:
Further proofs and discussion can be found in Naslund's and Bontea's solutions in this thread of the identical question!