Ytukyg

Let $f(x) = x^n-1$ be a polynomial in $Bbb R[x]$. Factorize $f(x)$ as a product of irreducible polynomials in $Bbb C[x]$ and show that if $n$ is even, $f(x)$ has two reals roots and if $n$ is odd, $f(x)$ has one real root.

Can someone help me? I know that a polynomium of degree $n$ has exactly $n$ complex roots and that I need to factorize it as a product of $n$ degree $1$ polynomials with $x-a_i$ where $a_i$ is the $i$'th root. But how do I determine the complex roots? And do I have to use the fact that complex roots always come in pairs with its complex conjugated?
I know that if $n $is even the are two real roots $1$ and $-1$ and if $n$ is odd the only root is $1$ with multiplicity of $2$.

Hint:

The complex roots are the $n$-th roots of unity, which you determine with the exponential form of complex numbers: they have modulus $1$ and their argument must satisfy
$$bigl(mathrm e^{itheta}bigr)^n=mathrm e^{intheta}=1=mathrm e^{icdot0}iff nthetaequiv 0mod 2piiffthetaequiv 0modfrac{2pi}n,$$
taking into account that the complex exponential function has period $2ipi$.

Can you proceed and show there are $n$ roots of unity and determine which pairs are conjugate?