Show that $\prod_{d|n}\Phi_{d}(x)=(x^{n}-1)$ Define for $n \geq 1$
$\Phi_{n}(x)=\prod_{\substack{1\leq k\leq n \\ gcd(k,n)=1}}(x-e^{\frac{2\pi i k}{n}})$.
Show that $\prod_{d|n}\Phi_{d}(x)=(x^{n}-1)$

Let 
$$\mathbb N_n=\{1,2,\ldots,n\}$$
and 
$$A_d=\{k\in \mathbb N_n\;|\; \gcd(k,n)=d\}$$
Hint 
Prove that $(A_d)_{d|n}$ is a partition of $\mathbb N_n$ and then
$$(x^n-1)=\prod_{k\in \mathbb N_n}(x-e^{2ik\pi/n})=\prod_{d|n}\prod_{k\in A_d}(x-e^{2ik\pi/n})$$
finally notice that $$k\in A_d\iff\gcd(k,n)=d\\\iff \gcd(k/d=k',n/d=d')=1\iff k'\in\{k\in \mathbb N_n\;|\; \gcd(k,d)=1\}$$
