Is the following a proof of Euler's product formula for sine (see (8) below)? If no, then why?
The sine can be a polynomial by the taylor series for sine, which coverage for all $x \in \mathbb{R}.$
As known, the roots of this polynomial are $n\pi,~n \in \mathbb{Z}$.
Therefore, for some constants $C_n$, we can write this polynomial like that: $$\sin(x) = \prod_{n=-\infty}^{\infty}C_n\left(x-\pi n\right) = C_0x\prod_{n=1}^{\infty}C_{-n}C_n\left(x-\pi n\right)\left(x+\pi n\right) \tag{1}$$ Let $B_n = C_{-n}C_n$ and $B_0 = C_0$. Therefore, $$\frac{\sin\left(x\right)}{x}=B_0\prod_{n=1}^{\infty}B_n\left(x^2-\left(\pi n\right)^2\right) \tag{2}$$
As known, $\lim_{x\to 0}{\frac{\sin\left(x\right)}{x}} = 1$. Hence, $$\lim_{x\to 0}{\left [B_0\prod_{n=1}^{\infty}B_n\left(x^2-\left(\pi n\right)^2\right) \right ]} = B_0\prod_{n=1}^{\infty}\left ( -B_n\left(\pi n\right)^2 \right ) = 1 \tag{3}$$
Therefore, $$B_0=\frac{1}{\prod_{n=1}^{\infty}\left(-B_n\left(\pi n\right)^2\right)} \tag{4}$$
Finally, $$\begin{align}\sin\left(x\right)&=B_0x\prod_{n=1}^{\infty}B_n\left(x^2-\left(\pi n\right)^2\right) \tag{5}\\ &=\frac{x\prod_{n=1}^{\infty}B_n\left(x^2-\left(\pi n\right)^2\right)}{\prod_{n=1}^{\infty}\left(-B_n\left(\pi n\right)^2\right)} \tag{6}\\ &=x\prod_{n=1}^{\infty}\frac{B_n\left(x^2-\left(\pi n\right)^2\right)}{\left(-B_n\left(\pi n\right)^2\right)} \tag{7}\\ &=x\prod_{n=1}^{\infty}\left(1-\left(\frac{x}{\pi n}\right)^2\right) \tag{8}\end{align}$$
Can this be a proof? Thank you.
