I am struggling with a proof of following identity $$ 1-z^m=\prod_{k=1}^m (1-\omega_kz)$$ where $\omega_k=e^{\frac{2\pi k}{m}i}$ for $k=1,2,\dotsc m$ are roots of unity and $z$ is a complex number. I've started with expanding RHS and I get $$1-z\sum_{j=1}^m \omega_j+z^2\sum_{\substack{ j=1 \\ k<j}}^m \omega_j\omega_k- \dots+(-1)^mz^m\prod_{j=1}^m\omega_j$$ Then I've proved that sum of roots of unity equals 0 and the product of roots of unity equals 1, but I don't know how to deal with those mixed sums.
I'll be grateful for any clues.
If $\omega$ is a root, then so is $\overline{\omega}$, except for $\omega=1$ and $\omega=-1$. $\omega=-1$ is only a root for even $m$, and as $\omega\overline{\omega}=1$, the product of the roots of unity is $(-1)^{m-1}$.
Next, it is fairly simple to prove that:
$$\sum_{j=1}^m \omega_j^n=0$$
for any $1\le n<m$, and $m>1$.
Then note that:
$$\left(\sum_{j=1}^m \omega_j\right)^2=\sum_{j=1}^m \omega_j^2+2\sum_{1\le i< j\le m} \omega_i\omega_j$$
so that $\sum_\limits{1\le i< j\le m} \omega_i\omega_j=0$.
The rest follow similarly, noting that:
$$\sum_{j=1}^m \omega_j^m=m$$