What is the five digit integer in the numerator of the simplified result?

Using the identity no. 40 on this page, we have,

$\left(1-\frac{1}{3^4}\right)\left(1-\frac{1}{6^4}\right)\prod_{n=3}^{\infty}\left(1-\frac{1}{\left(3n\right)^4}\right)=\frac{3^3\sin\left(\frac{\pi}{3}\right)}{\pi\left|\left(-\frac{i}{3}-1\right)!\right|^2}$

Because $\displaystyle z=\frac{1}{3}$ . And so, after simplification, we have

$\prod_{n=3}^{\infty}\left(1-\frac{1}{\left(3n\right)^4}\right)=\frac{59049\sqrt{3}\sinh\left(\frac{\pi}{3}\right)}{12950\pi^2}$

A Simpler Approach :-

Using the identity

$\prod_{n=3}^{\infty}\left(1-\frac{1}{\left(3n\right)^4}\right)=\prod_{n=3}^{\infty}\left(1-\frac{1}{\left(3n\right)^2}\right)\cdot\prod_{n=3}^{\infty}\left(1+\frac{1}{\left(3n\right)^2}\right)$

we can evaluate the products on the right hand side using the infinite product for the Sinc function i.e.

$\frac{\sin x}{x}=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right)$

Putting $\displaystyle x=\frac{\pi}{3}$ and $\displaystyle x=\frac{i\pi}{3}$ in the above infinite product, and after some simplifications, we arrive at the same answer.

$\prod _{n=3}^{\infty } \left(1-\frac{16}{(6 n)^4}\right) \Rightarrow \frac{59049 \sqrt{3} \sinh \left(\frac{\pi }{3}\right)}{12950 \pi ^2}$ .