Lower bound on divisors of $\Phi_n(n) $

Question

Take the nth cyclotomic polynomial $\Phi_n(x)$ and let $\phi$ be the Euler totient function. I can prove that all divisors $d$ of $\Phi_n(n)$ are such that $d \ge \phi(n)$ or $d = 1$.

The proof is not that intuitive and involves algebraic number theory. I'm wondering if there is an intuitive elementary proof of this fact.

Example $n=5$:
$\Phi_5(5) = 781 = 11 \times 71$
$\phi(5) = 4$ and indeed both $11, 71$ are larger than $4$

Answer 1

I'm not sure if you will find this proof intuitive, but it is certainly elementary.

Cyclotomic polynomials have the following (elementary) property:

If $n>0$ and $a\in\mathbb Z$, then every prime divisor $p$ of $\Phi_n(a)$ satisfies either $p\mid n$ or $p\equiv1\pmod n$.

A proof can be found for example here [Theorem 6].

In our case $\Phi_n(n)\mid n^n-1$, so if $p\mid\Phi_n(n)$ we certainly have that $p\nmid n$.

From the property above it follows that every prime divisor of $\Phi_n(n)$ is at least $n+1$, which is bigger then $\phi(n)$.