For $n \in \mathbb{N}$, we define $\Phi_n \in \mathbb C[x]$ as the monic polynomial that has as roots the $n$th primitive roots of the unity.
For example $\Phi_2 =(x+1)$,
$\Phi_4 = (x-i)(x+i) = x^2+1$
Prove that if $k|n$ and $0<k<n$ then $$\Phi_n | \frac {x^n-1}{x^k-1} $$
Let $z_0 \in \Bbb C$ be a primitive $n^{th}$ root of unity. So,$z_0^n=1$ and $z_0^k \neq 1 $ since $k<n$.Let $P(x)=\frac{x^n-1}{x^k-1}$, $$P(z_0)=\frac{z_0^n-1}{z_0^k-1}=0$$ $$P(x)=\frac{x^n-1}{x^k-1}=x^{k(l-1)}+x^{k(l-2)}+x^{k(l-3)}+....+1$$
So,$P$ is a polynomial and $\Phi | P$