Proof of the product formula for sine function I am looking for a simple way to prove $$\frac{\sin \pi z}{\pi z}=\prod_{n=1}^\infty \left(1-\frac{z^2}{n^2}\right)$$ using mainly on the fact that the entire function has simple zeros at $n=\pm 1, \pm 2,\cdots$.
 A: I know one proof that does not involve infinite product.
Think about the equation
$\sum_{n = 1}^\infty \frac{2z}{n^2\pi^2 - z^2} = -\cot z + \frac{1}{z}$,
and integrate both sides. For a proof of this equation, consider the function $f(z) = \frac{\cot z}{z(z - 1)}$, then use calculus on residues.
A: We will use Hardamard Factoriztation theorem to prove it (see https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem#Hadamard_factorization_theorem).
Observe $|\sin(\pi z)|\le e^{\pi |z|}$,  hence it has order of growth less than equal to one. Further it has simple zero at every integer $n \in Z$. Hence by Hardamard Factorization theorem we get
$$ \sin (\pi z)= z e^{az+b}\displaystyle \prod_{0 \ne n \in Z}\left(1-\frac zn\right)e^{\frac z n}, \text { for some } a, b \in  C.$$ Pairing $n$ and $-n$ together in the product  we get
\begin{equation}
 \sin (\pi z)= z e^{az+b}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).
\label{on}
\end{equation}
From above equation we will get
$$\frac {\sin (\pi z)}{\pi z}= \frac{e^{az+b}}{\pi}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
Taking $z \to 0$, in both side we get $\frac{e^{b}}{\pi}=1$.
From above equation we will get
$$\frac {\sin (\pi z)}{\pi z}= {e^{az}}\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$ Again using the fact that
$\frac {\sin (\pi z)}{\pi z}$ is an even function, it can be easily seen that $a=0$.
Hence we get $$\frac {\sin (\pi z)}{\pi z}= \displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
Thus  $${\sin (\pi z)}= \pi z\displaystyle \prod_{n =1}^{\infty}\left(1-\frac {z^2}{n^2}\right).$$
