Here is another answer that uses the basic definition
$$\Phi_n(x) = \prod_{1\le k\le n \atop \gcd(k,n)=1} (x-e^{2\pi i k/n}).$$
This gives
$$\Phi_n(0) = (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1} e^{2\pi i k/n}.$$
Now recall that for $n>1$ we have $$\sum_{1\le k\le n \atop \gcd(k,n)=1} k = \frac{1}{2} \varphi(n) n.$$
This gives
$$\Phi_n(0) = (-1)^{\varphi(n)} e^{(2\pi i/n) \times 1/2\times \varphi(n) \times n}
= (-1)^{\varphi(n)} e^{\pi i \varphi(n)} = (-1)^{2\varphi(n)} =1.$$
For the second part we get that
$$\Phi_n(-x) = \prod_{1\le k\le n \atop \gcd(k,n)=1} (-x-e^{2\pi i k/n})
= (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1}
(x+e^{2\pi i k/n}) \\ =
(-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1}
(x-e^{\pi i + 2\pi i k/n})=
(-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1}
(x-e^{2\pi i (k/n+1/2)})\\
= (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1}
(x-e^{2\pi i (2k+n)/(2n)}).$$
This almost concludes the proof. We can ignore the sign because $\varphi(n)$ is even.
We now claim there is a bijection between the primitive roots modulo $2n$ call them of class $\mathcal{P}_1$ given by $e^{2\pi i k/(2n)}$ where $\gcd(k,2n)=1$ and $1\le k\le 2n$ and the values produced by $e^{2\pi i (2k+n)/(2n)}$ call them of class $\mathcal{P}_2$ where $\gcd(k,n)=1$ and $1\le k\le n.$ (These two conditions will be our domain of $k.$)
Let $\mathcal{Q}$ be the class of primitive roots modulo $n$. The $\mathcal{P}_1$ roots can be obtained from these by taking $q$ as is if it is odd and taking $q+n$ if it is even (recall that $n$ is odd). Now to establish a bijection to $\mathcal{P}_2$ we must show that $q\equiv 2k+n\bmod (2n)$ and $q+n\equiv 2k+n\bmod (2n)$ always have a unique solution $k$ in the domain of $k$. In the first case we take $k=(q+n)/2$ which is easily seen to be in this domain. In the second case take $k=q/2,$ also clearly in this domain. Moreover we clearly hit each value of $k$ exactly once. This finishes the proof.