Prove that the roots of cyclotomic polynomial $\Phi_{p-1}(x) \equiv 0 (mod~p)$ are exactly the primitive roots mod p

Question

$p$ is a prime, and $\Phi_{p-1}(x)$ denote the cyclotomic polynomial of order $p-1$. And I want to show the following:

$g$ is a solution of the congruence $\Phi_{p-1}(x) \equiv 0 (mod~p)$ if and only if $g$ is a primitive root (mod p)

that is:

$\Phi_{p-1}(g) \equiv 0 (mod~p) \iff g$ is a primitive root (mod p)

Here is some properties about cyclotomic polynomial:

${\textstyle \prod_{d|n}^{}}\Phi_{d}(x) = x^n-1\tag{1}$

$\Phi_{n}(x) = {\textstyle \prod_{d|n}^{}}(x^d-1)^{\mu(n/d)}\tag{2}$

$\mu(x)$ is the Möbius inversion formula