<s>[INST] Infinite Product $\prod\limits_{k=1}^\infty\left({1-\frac{x^2}{k^2\pi^2}}\right)$ I've been looking at proofs of Euler's sine expansion, that is 
$$\frac{\sin x}{x}=\prod_{k=1}^\infty\left({1-\frac{x^2}{k^2\pi^2}}\right)$$
All the proofs seem to rely on complex analysis and Fourier series. 
Is there any more elementary proof?
 [/INST] I first prove that for any $n$ that is a power of $2$ and for any $x\in\mathbb R$,
$$\sin(x)=n p_n(x)\sin\left(\frac x n\right)\cos\left(\frac x n\right),~p_n(x)=\prod_{k=1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\tag{1}$$
using the following more familiar identity
$$\sin(x)=2\sin\frac x 2\sin\frac{\pi+x}2$$
we can easily arrive to the following identity that is valid for $n$ equal to any power of $2$,
$$\begin{align}\sin(x)&=2^{n-1}\prod_{k=0}^{n-1}\sin\frac{k\pi+x}n\\
&=2^{n-1}\sin\frac x n\sin\frac{\frac n 2\pi+x}n\prod_{k=1}^{\frac n 2-1}\sin\frac{k\pi+x}n\sin\frac{k\pi-x}n\\
&=2^{n-1}\sin\frac x n\cos\frac x n\prod_{k=1}^{\frac n 2-1}\left(\sin^2\frac{k\pi}n-\sin^2\frac x n\right)\tag{2}\end{align}$$
By considering what happens as $x\to0$ in the recent formula, we can obtain that for $n$ a power of $2$
$$n=2^{n-1}\prod_{k=1}^{\frac n 2-1}\sin^2\frac{k\pi}n\tag{3}$$
and then I replace $(3)$ in $(2)$ and prove $(1)$.
Now I choose $m<\frac n 2-1$ and break up $p_n(x)$ as follows
$$p_n(x)=\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\tag{4}$$
Since for $0<\theta<\frac{\pi}2$, $\sin(\theta)>\frac2{\pi}\theta$, we have
$$\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<\frac{x^2}{4k^2}$$
and if I choose $m$ and $n$ large enough such that $\frac{x^2}{4m^2}<1$, then
$$0<1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<1,k=m+1,...,\frac n 2-1$$
which implies that
$$0<\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)<1$$
Therefore from $(4)$ we have
$$p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)$$
Also we have
$$\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\ge1-\sum_{k=m+1}^{\frac n 2-1}\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\ge1-\frac{x^2}4\sum_{k=m+1}^\infty\frac1{k^2}=1-s_m$$
and once again by $(4)$, we get
$$p_n(x)\ge(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)$$
To summarize, I have shown that
$$(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\le p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\tag{5}$$
Now taking $n\to\infty$ in $(5)$ I arrive to
$$(1-s_m)\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\le\frac{\sin(x)}x\le\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)$$
After rearrangement and then taking absolute values, we get
$$\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|$$
but since we have
$$\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le\prod_{k=1}^m\left(1+\frac{x^2}{k^2\pi^2}\right)\le\prod_{k=1}^me^{\frac{x^2}{k^2\pi^2}}\le e^{\sum_{k=1}^\infty\frac{x^2}{k^2\pi^2}}=e^{\ell}$$
I can write
$$\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|e^{\ell}$$
Finally since $\lim_{m\to\infty}s_m=0$, this inequality implies our goal, indeed
$$\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)=\frac{\sin(x)}x$$
</s>