Show that $\displaystyle X^n-1=\prod_{d\mid n}\Phi_d(X)$.
We have that $$\Phi_n(X)=\prod_{\underset{\gcd(i,n)=1}{1\leq i\leq n}}(X-\zeta_n^i)$$ where $\zeta_n=e^{\frac{2i\pi}{n}}$ therefore, we have to show that $$X^n-1=\prod_{d\mid n}\prod_{\underset{\gcd(i,d)=1}{1\leq i\leq d}}(X-\zeta_d^i).$$
My first idea is to show that both has same roots and that all their roots are distinct. For the unicity of the roots of $X^n-1$ it's obvious. It's ovbious to that all roots of $\Phi_d(X)$. But how can I show that they have the same roots and that the roots of $$\prod_{d\mid n}\Phi_d(X)$$ are distinct ?
$$X^n-1=\prod_{d\mid n}\prod_{\underset{\gcd(i,d)=1}{1\leq i\leq d}}(X-\zeta_d^i).$$
You need to show that
$$\prod_{\underset{\gcd(i,d)=1}{1\leq i\leq d}}(X-\zeta_d^i)=\Phi_d(X)$$
Consider the RHS, its roots are the primitive $d^{th}$ roots of unity i.e. if $\zeta$ is a root then $|\zeta|=d$. Clearly LHS is satisfied by all such $\zeta$'s since the LHS is satisfied by all the $n^{th}$ roots of order $d$ (Since, $|\zeta_d^i|=d$) and $RHS|LHS$
Also degree of LHS and RHS is same (thus, the $\phi(d)$ distinct roots of $\Phi_d(X)$ satisfy LHS) and both are monic and thus we can conclude that LHS=RHS and your claim is proved.