Does this the sequence go to zero?

$\Pi_{n=1}^{N}\text{sin}(2\pi n\omega)$ as N $\rightarrow \infty$ for any $\omega \in (0,1)?$

I can see this sequence is always decreasing for general $\omega$. And for some specific $\omega$ (in $\mathbb{Q}$ I believe), sin($2\pi n_0\omega$) becomes 0 for some specific $n_0$ so making the sequence to be 0 there after. But how to show such sequence goes to zero for general $\omega \in (0,1)$ though?