Infinite product of sine function [closed]

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?

closed as off-topic by Gerald Edgar, Noah Schweber, Myshkin, Alexey Ustinov, Todd Trimble♦Oct 29 '15 at 5:23

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• See "help" above to find what sort of questions to ask here. – Gerald Edgar Oct 29 '15 at 2:43
• Hint: Kronecker's theorem for $\omega\not\in\mathbb Q$. – Fan Zheng Oct 29 '15 at 3:12
• Fan Zheng - what theorem do you mean exactly? – Sergei Oct 29 '15 at 12:32