Show that if p is an odd prime dividing Φd(a) then either p|d or p≡1 mod d."

Question

Let a ∈ Z, d ∈ N. Consider the dth cyclotomic polynomial, Φ_d(x). Show that if p is an odd prime dividing Φ_d(a) then either $p|d$ or $p ≡ 1$ mod d."

I am given the solution below; but my question is, how does showing $D=d$, and the order of a mod p is d, prove that what we need to show, that "either $p|d$ or $p ≡ 1$ mod d"? I am not sure I follow the general proof strategy here.

---

Here is the solution: Suppose p does not divide d. Since Φ_d(x)|(x^d − 1), $$Φ_d(a) ≡ 0 \mathrm{\;mod\;p} ⇒ a^d − 1 ≡ 0 \mathrm {\;mod\;p}$$ so a is coprime to p and the order of a mod p is a divisor of d.

Now suppose that the order of a mod p is a divisor, D of d, with $D < d$.

Then \begin{equation*} a^D-1 = \prod_{e|D} Φ_e(a) ≡ 0 \mathrm {\;mod\;p} \end{equation*} . So there exists $e|D$ such that $Φ_e(a) ≡ 0$ mod p, and $e < d$ since $e|D$ and $D < d$ by assumption. It follows that a is a root of \begin{equation*} x^d − 1 = \prod_{f|d} Φ_f(a) ≡ 0 \mathrm {\;mod\;p} \end{equation*}

with multiplicity at least 2, i.e. that $$x^d − 1 ≡ (x − a)^2 \cdot f(x) \mathrm {\;mod\;p}$$ for some polynomial f(x) ∈ Z[x]. But this is impossible, since after taking derivatives, we would get that a must also be a root of $dx^d$ mod p , but $da^d \not\equiv 0$ mod p, since p does not divide d and p does not divide a. So $D = d$ and hence the order of a mod p is exactly d.