I want to prove the formula: $\sin(x)=x\prod\limits_{k=1}^\infty\left( 1-\cfrac{x^2}{\pi^2k^2}\right)$
I found a video with a proof here https://www.youtube.com/watch?v=WHGx3gu1QiM
but is incomplete.
I inspired from there and I proved by induction this: $\sin(x)=2^k\sin\left(\cfrac{x}{2^k}\right)\cos\left(\cfrac{x}{2^k}\right)\left( 1-\cfrac{\sin^2(\frac{x}{2^k})}{\sin^2(\frac{\pi}{2^k})}\right)...\left( 1-\cfrac{\sin^2(\frac{x}{2^k})}{\sin^2(\frac{(2^{k-1}-1)\pi}{2^k})}\right),\forall k\in \mathbb{N_{\geq2}} (*)$
Now if $k\rightarrow\infty$ in $(*)$ then "almost" we get: $\sin(x)=x\prod\limits_{k=1}^\infty\left( 1-\cfrac{x^2}{\pi^2k^2}\right)$ but we must prove something like this $$\lim\limits_{k\to\infty} f_1(k)f_2(k)...f_{\phi(k)}(k)=\prod\limits_{n=1}^\infty \lim\limits_{k\to\infty}f_n(k) $$ ,where $\phi(k)=2^{k-1}+2$(number of elemnts in $(*)$) in our particular case.
My question is: How I can prove the fomula $\lim\limits_{k\to\infty} f_1(k)f_2(k)...f_{\phi(k)}(k)=\prod\limits_{n=1}^\infty \lim\limits_{k\to\infty}f_n(k) $? is there a theorem in mathematics?