Cyclotomic Polynomials and Euler's Totient Function

Question

Claim: $$\prod_{d \mid n} \Phi_{d}(x) = x^n - 1 $$ Where $\Phi_{m}(x)$ is the $m^{th}$ cyclotomic polynomial of $x$.

I think it has to do with Euler's Totient Function $\phi$ and the result $$\sum_{d \mid n} \phi(d) = n$$ but am not sure how to show the intermediate step(s).

Edit:

Ok, I would now like to take this a little further and show, by induction on $m$ and use of the above result, that $$\Phi_m (x)\in \mathbb{Z}[x]$$ By basic computation I know it's true for $m<4$. I've considered dividing the products from the initial claim for $n=k$ and $n=k+1$. Do you have any advice?

Answer 1

The roots of $x^n-1$ are the $n$-th roots of unity, $\omega_n^k$ for $k=0,\dotsc,n-1$ with $\omega_n=\mathrm e^{2\pi\mathrm i/n}$. The roots of the cyclotomic polynomial $\Phi_d(x)$ are the primitive $d$-th roots of unity. Every $n$-th root of unity is a primitive $d$-th root of unity for exactly one divisor $d$ of $n$. The result follows.

Answer 2

You can do this by looking at the roots of $f(x)=x^n-1$, which come out as $e^{\frac {2\pi i}{n}}$, and which (multiplicatively) form a cyclic group of order n.

The cyclotomic polynomials pick out the roots having exact order $d$ for each $d|n$ (the primitive roots mod $d$).

Each element of the group has an order which is a divisor of $n$ and is therefore captured by the relevant cyclotomic polynomial.

There are $\phi(d)$ primitive roots mod $d$, so the order of each $\Phi(d)$ is $\phi(d)$.