I don't understand how to prove $X^n -1 = \prod_{d|n}{\Phi_d}$ where $\Phi$ is the cyclotomic polynomial

Question

I want to show that $X^n-1 = \prod_{d|n}{\Phi_d}$, and I found this proof on internet:

$X^n-1=\prod_{k=1}^{n}{(X-e^{i\frac{2k\pi}{n}})}=\prod_{d|n}{\prod_{k=1,gcd(k;n)=d}^{n}{(X-e^{i\frac{2k\pi}{n}})}}=\prod_{d|n}{\Phi_{n/d}}=\prod_{d|n}{\Phi_d}$

What I don't understand :

1) How to split the product in a double one (right hand side of the second equality) and how to justify it.

2) How to pass from $\Phi_{n/d}$ at $\Phi_d$ at the end? Is it beacause there is a bijection $D(n) -> D(n); d -> n/d$ (where $D(n) = \left\{d|n , d\ge0\right\}$)

Answer 1

1) We can write the set $\{1,\cdots,n\}$ as the disjoint union of the sets $\{k;\,1\le k\le n\,\mathrm{and}\,gcd(k,n)=d\}$ for all $d$ belonging to the set of (positive) divisors of $n$.

2) Yes, the bijection you mention is the main key (another reason lies in the fact that multiplication of integers is a commutative law).

Answer 2

Since the $n$-th roots of unity form a cyclic group, $$ X^n-1=\prod_{ord(\omega) \mid n}{X-\omega} =\prod_{d \mid n} \prod_{ord(\omega)=d}{X-\omega} =\prod_{d \mid n} \Phi_d(X) $$ where $ord(\omega)$ is the least positive exponent $k$ such that $\omega^k=1$.