The cyclotomic polynomials arise from a very simple equation: the equation $z^n=1$, whose complex roots, the $n$th roots of unity, are given by $1,e^{\frac{2\pi i}{n}},...,e^{\frac{2\pi i(n-1)}{n}}$, where $e^{ix}=\cos x +i\sin x$. The number $n$ will always be a positive integer. The roots form a regular $n$-gon in the complex plane. A root $\zeta$ of the equation $z^n=1$ is called a primitive $n$th root of unity if $\zeta^k\neq 1$ for $k<n$, or equivalently if the set $\{1,\zeta,...,\zeta^{n-1}\}$ contains all the roots of $z^n=1$. Then we define the $n$th cyclotomic polynomial as
\[\begin{eqnarray}\Phi_n(x):=\prod_{\zeta\in S_n}(x-\zeta),\end{eqnarray}\]
where $S_n$ is the set of primitive $n$th roots of unity. Clearly the primitive roots are $e^{\frac{2\pi i k}{n}},$ where $(k,n)=1$, so there are $\varphi(n)$ of them. Hence $\Phi_n(x)$ is a monic polynomial of degree $\varphi(n)$. If $\zeta$ is any $n$th root of unity, then $\zeta$ is in a unique $S_d$ where $d\leq n$. We have $d\mid n$ since if $n=kd+r$ with $0<r<d$, we would actually have $\zeta\in S_r$. This gives the important identity
\[\begin{eqnarray}x^n-1=\prod_{d\mid n}\Phi_d(x),\end{eqnarray}\]
since both sides are monic polynomials and the roots of the right-hand side are the elements of the sets $S_d$ for $d\mid n$ (there are no multiple roots as each root belongs to a unique $S_d$). This formula allows us for example to calculate cyclotomic polynomials explicitly. Obviously $\Phi_1(x)=x-1$, and then
\[\begin{eqnarray}\Phi_n(x)=\frac{x^n-1}{\prod'_{d\mid n} \Phi_d(x)},\end{eqnarray}\]
\[\begin{eqnarray}\Phi_n(x)=\frac{x^n-1}{\prod'_{d\mid n} \Phi_d(x)},\end{eqnarray}\]
where $'$ denotes that the term $d=n$ is ignored. Hence each cyclotomic polynomial is determined by the previous ones. In particular, the following holds.