Proof

Note that $\omega^n = 1$, suppose $n = 2k + 1$ then

$$\begin{align} \prod_{i=0}^{2k} \omega^i & = \prod_{i=1}^{2k} \omega^i\\ & = \prod_{i=1}^{k} \omega^i \omega^{2k + 1 - i}\\ & = \prod_{i=1}^k \omega^{2k+1}\\ & = 1\end{align}$$

this proves the case for odd $n$. Now instead suppose $n = 2k$ then

$$\begin{align} \prod_{i=0}^{2k-1} \omega^i & = \prod_{i=1}^{2k -1} \omega^i\\ & = \omega^k \prod_{i=1}^{k-1} \omega^i \omega^{2k - i}\\ & = - \prod_{i=1}^{k-1} \omega^{2k}\\ & = - 1\end{align}$$

which proves the case for even $n$.