I have thought for hours but got no clue how to get this infinite product
Once these expressions are substituted into $(2.5)$, all the real terms in the brackets (the terms in even powers of $x$) cancel out. As soon as the common factor of $2ix$ is factored out in the remaining odd-power terms of $x$, the right-hand side of $(2.5)$ reduces to a compact form, and we have $$\sin{x}=x\lim_{n\to \infty} \sum_{k=0}^{(n-1)/2} (-1)^k \binom{2k+1}{n} \frac{x^{2k}}{n^{2k+1}} \tag{2.8}$$ Of all the stages in Euler's procedure, which, as a whole, represents a real work of art, the next stage is perhaps the most critical and decisive. Factoring the polynomial in $(2.8)$ into the trigonometric form $$\sin{x}=x\lim_{n\to \infty} \prod_{k=1}^{(n-1)/2}\left[1-\frac{(1+\cos 2k\pi/n)x^2}{(1-\cos 2k\pi/n)n^2}\right]$$
An excerpt from the book 'Greens functions and infinite Products' written by Melnikov Yu.
