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'Hôpital's rule for the limit of the ratio as $x\to1$),
and the denominator gives you
an inductive statement about
$$a_n=\begin{cases}1,&n=1;\\\Phi_n(1),&n>1,\end{cases}$$
namely
$$
a_1=1,\quad
a_n=\frac{n}{\prod_{1<d<n,~d|n}a_d}
$$
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 $n=\prod_{0<d|n}a(n)$
(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 $f,g$:
$$
\log(a(n))
=\sum_{0<d|n}\mu\left(\frac{n}{d}\right)\log d
=\Lambda(n),
$$
where $\Lambda(n)$ is the von Mangoldt function, so
\begin{align}
a(n)&
=\prod_{0<d|n}\mu\left(\frac{n}{d}\right)\log d
=e^{\Lambda(n)}\\
&=n^{\epsilon(\omega(n))/\Omega(n)}
=\begin{cases}p&\text{if }n=p^k\text{ is a power of a single prime};
\\1&\text{otherwise, i.e. if }n=1\text{ or }\omega(n)>1,\end{cases}
\end{align}
where $\omega(n)$ and $\Omega(n)$ are the prime omega functions, so that
$$
\Phi_n(1)=\begin{cases}
0&\text{if }n=1;\\
p&\text{if }n=p^k;\\
1&\omega(n)>1.\end{cases}
$$
EDIT:
Further proofs and discussion can be found in Naslund's and Bontea's solutions in this thread of the identical question!
