Understanding a particular evaluation of $\prod\limits_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right)$ I'm having a hard time understanding the following evaluation of the infinite product  $$ \prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right).$$
In particular, I don't understand how you go from line 2 to line 3.
Here $\omega = -\frac{1}{2}+ i \frac{\sqrt{3}}{2}$, which is a primitive third root of unity.

$$\begin{align}\prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right)  &=\prod_{n=2}^{\infty}\frac{(n-1)(n^2+n+1)}{n^3} \\  &=\lim_{m\to\infty}\frac{1}{m}\prod_{n=2}^m\frac{(n-\omega)(n-\omega^2)}{n^2} \\  &=\lim_{m\to\infty}\frac{\Gamma(m+1-\omega)\Gamma(m+1-\omega^2)}{m(m!)^2\Gamma(-\omega)\Gamma(-\omega^2)(1-\omega)(1-\omega^2)(-\omega)(-\omega^2)} \\  &=\frac{1}{3\Gamma(-\omega)\Gamma(-\omega^2)} \\ &=\frac{\sin{\pi(-\omega)}}{3\pi} \\  &=\frac{\cosh (\frac{\sqrt{3}\pi}{2})}{3\pi } \end{align}$$
EDIT: I think I'm starting to make sense out of this by writing out the terms.
$$ \begin{align} \prod_{n=2}^{m} \frac{(n- \omega)(n-\omega^{2})}{n^{2}} &= \frac{(2-\omega)(3-\omega) \cdots (m- \omega)(2-\omega^{2})(3-\omega^{2})\cdots (m- \omega^{2})}{2^{2} \cdot 3^{2} \cdot \cdots  \cdot m^{2}}  \\ &= \frac{1}{(m!)^{2}} \frac{\Gamma(m+1-\omega)}{(1-\omega)(-\omega)\Gamma(-\omega)} \frac{\Gamma(m+1-\omega^{2})}{(1-\omega^{2})(-\omega^{2})\Gamma(-\omega^{2})} \end{align}$$
Now I have to figure out how to take the limit as $m \to \infty$.
 A: Here's my understanding of that evaluation.
$$  \begin{align} \prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right) &= \prod_{n=2}^{\infty}\frac{(n-1)(n^2+n+1)}{n^3} \\ & = \prod_{n=2}^{\infty} \frac{(n-1)(n- \omega)(n-\omega^{2})}{n^{3}} \\ &= \lim_{m \to \infty} \prod_{n=2}^{m} \frac{n-1}{n} \frac{(n- \omega)(n-\omega^{2})}{n^{2}}  \\ &= \lim_{m \to \infty}  \frac{1}{m}\frac{(2-\omega)(3-\omega) \cdots (m- \omega)  (2-\omega^{2})(3-\omega^{2})\cdots (m- \omega^{2})}{2^{2} \cdot 3^{2} \cdot \cdots  \cdot m^{2}}  \\ &= \lim_{m \to \infty} \frac{1}{m} \frac{1}{(m!)^{2}}  \frac{\Gamma(m+1-\omega)}{(1-\omega)(-\omega)\Gamma(-\omega)} \frac{\Gamma(m+1-\omega^{2})}{(1-\omega^{2})(-\omega^{2})\Gamma(-\omega^{2})} \\ & \stackrel{(1)}= \frac{1}{3\Gamma(-\omega) \Gamma(-\omega^{2})}\lim_{m \to \infty} \frac{\Gamma(m+1- \omega) \Gamma(m+1-\omega^{2})}{m \Gamma(m+1) \Gamma(m+1) }  \\ & \stackrel{(2)}=  \frac{1}{3\Gamma(-\omega) \Gamma(-\omega^{2})}\lim_{m \to \infty} \frac{\Gamma(m+1- \omega) \Gamma(m+1-\omega^{2})}{m^{-\omega}m^{-\omega^{2}} \Gamma(m+1) \Gamma(m+1) } \\ &=   \frac{1}{3\Gamma(-\omega) \Gamma(-\omega^{2})}\lim_{m \to \infty} \frac{(m-\omega)(m-\omega^{2})}{m^{2}}\frac{\Gamma(m- \omega)}{\Gamma(m) m^{-\omega} } \frac{\Gamma(m-\omega^{2})}{ \Gamma(m)m^{-\omega^{2}} } \\ & \stackrel{(3)}=  \frac{1}{3\Gamma(-\omega) \Gamma(-\omega^{2})} (1)(1)(1) \\ & = \frac{1}{3 \Gamma \left(\frac{1}{2} - i\frac{\sqrt{3}}{2} \right) \Gamma \left(\frac{1}{2} + i\frac{\sqrt{3}}{2} \right)} \\& = \frac{1}{3 \Gamma \left(1- \frac{1+ i \sqrt{3}}{2} \right) \Gamma \left(\frac{1}{2} + i\frac{\sqrt{3}}{2} \right)}  \\  & \stackrel{(4)}= \frac{\sin \left(\pi \, \frac{1+i \sqrt{3}}{2}  \right)}{3 \pi} \\&= \frac{\sin \left( \frac{\pi}{2}\right) \cos \left(\frac{i \sqrt{3} \pi}{2} \right) + \cos \left( \frac{\pi}{2} \right) \sin \left(\frac{i  \sqrt{3} \pi }{2}\right)}{3 \pi} \\ &= \frac{\cosh \left( \frac{\sqrt{3} \pi}{2} \right)}{3 \pi} \end{align}$$

$(1)$ $(1-\omega)(1-\omega^{2}) = 3$ and $(-\omega)(-\omega^{2})=1 $
$(2)$ $-\omega - \omega^{2} =1$
$(3)$ $\lim_{n \to \infty} \frac{\Gamma(n+ \alpha)}{\Gamma(n)n^{\alpha}}=1 $
$(4)$ Reflection formula for the gamma function
