<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 think this is the most motivated elementary proof.
For $z \in \mathbb C$ and $n \in \mathbb N_{> 0}$, let:
$$f_n \left({z}\right) = \frac 1 2 \left[{\left({1 + \frac z n}\right)^n - \left({1 - \frac z n}\right)^n }\right]$$
Then $f_n\left({z}\right) = 0$ if and only if:
\begin{align}
&&\left({1 + \frac z n}\right)^n & = \left({1 - \frac z n}\right)^n \\
& \iff & 1 + \frac z n & = \left({1 - \frac z n}\right) e^{2 \pi i \frac k n} \\
& \iff & z & =  n \frac {e^{2 \pi i \frac k n} - 1} {e^{2 \pi i \frac k n} + 1} \\
&&& =  n i \tan \left({\frac {k \pi} n }\right)
\end{align}
Let $n = 2 m + 1$.
Then the roots of $f_{2 m + 1} \left({z}\right)$ are $\left({2 m + 1}\right) i \tan \left({\dfrac {k \pi} {2 m + 1}}\right)$ for $- m \le k \le m$.
Observe that $f_{2m + 1} \left({z}\right)$ is a polynomial of degree $2 m + 1$. 
Then for some constant $C$, we have:
\begin{align}
f_{2 m + 1} \left({z}\right) & = C z \prod_{\substack {k \mathop = - m \\ k \mathop \ne 0} }^m \left({1 - \frac z {\left({2 m + 1}\right) i \tan \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)\\
& = C z \prod_{k \mathop = 1}^m \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)
\end{align}
It can be seen from the binomial theorem that the coefficient of $z$ in $f_n \left({z}\right)$ is $1$.
Hence $C = 1$, and we obtain:
$$f_{2 m + 1} \left({z}\right) = z \prod_{k \mathop = 1}^m \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)$$
First we consider $z = x$ where $x$ is a non-negative real number.
Let $l < m$.
Then:
$$x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) \le f_{2 m + 1} \left({x}\right)$$
Taking the limit as $m \to \infty$ we have:
\begin{align}
& &\lim_{m \to \infty} x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} \left({\frac {k \pi / \left({2 m + 1}\right)} {\tan \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)^2 }\right) & \le \frac 1 2 \left({e^x - e^{- x} }\right)\\
& \implies &x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} }\right) & \le \sinh x
\end{align}
By the inequality $\tan \theta \ge \theta$ for $0 \le  \theta < \dfrac {\pi} 2$ we have:
$$f_{2 l + 1} \left({x}\right) \le x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} }\right) \le \sinh x$$
Taking the limit as $l \to \infty$ we have by Squeeze Theorem:
$$\quad x \prod_{k \mathop = 1}^\infty \left({1 + \frac {x^2} {k^2 \pi^2} }\right) = \sinh x \tag{1}$$
Now let $1 < l < m$.
We have:
\begin{align}
&\left \vert{f_{2 m  + 1} \left({z}\right) - z \prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)}\right \vert \\
& =\left \vert{z}\right \vert \left \vert{\prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)}\right \vert \cdot \left \vert{\prod_{k \mathop = l + 1}^m \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) - 1}\right \vert\\
& \le \left \vert{z}\right \vert \left[{\prod_{k \mathop = 1}^l \left({1 + \frac {\left \vert{z}\right \vert^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)}\right] \cdot \left[{\prod_{k \mathop = l + 1}^m \left({1 + \frac {\left \vert{z}\right \vert^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) - 1}\right] \\
& = f_{2 m  + 1} \left({\left \vert{z}\right \vert}\right) - \left \vert{z}\right \vert \prod_{k \mathop = 1}^l \left({1 + \frac {\left \vert{z}\right \vert^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)
\end{align}
Taking the limit as $m \to \infty$ we have:
$$\left \vert{\sinh z - z \prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {k^2 \pi^2} }\right)}\right \vert \le \sinh {\left \vert{z}\right \vert} - \left \vert{z}\right \vert \prod_{k \mathop = 1}^l \left({1 + \frac {\left \vert{z}\right \vert^2} {k^2 \pi^2} }\right)$$
Now take the limit as $l \to \infty$.
By $(1)$ and Squeeze Theorem, we have:
$$\sinh z = z \prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {k^2 \pi^2} }\right)$$
Finally, substituting $z \mapsto i z$, we obtain:
$$ \sin z = z \prod_{k \mathop = 1}^l \left({1 - \frac {z^2} {k^2 \pi^2} }\right)$$
</s>