Hadamard production for $\sin z$. We know that all zeros of $\sin z$ are $z=k\pi,\,k\in\mathbb Z$, all of which are of order $1$.

So by Hadamard Factorization theorem, we have
$$
\sin z=e^{P(z)}\cdot z\cdot\prod_{n=1}^{\infty}(1-\frac{z^2}{n^2\pi^2}),
$$
where $P(z)$ is a polynomial of degree $\leq1$.

Actually, 
$$
\sin z=z\cdot\prod_{n=1}^{\infty}(1-\frac{z^2}{n^2\pi^2}).
$$

So I wonder how to prove $P(z)=0$?
You already know that $P$ has degree at most one: $P(z) = az+b$. Then $\sin(-z) = -\sin(z)$ implies $e^{P(z)} = e^{P(-z)}$, i.e. $1 = e^{P(z)-P(-z)} = e^{2az}$, so that $a=0$  and $P$ is constant.

Then use $\frac{\sin(z)}{z} \to 1$ for $z\to0$ to conclude that $e^{P(z)} = 1$.
You already know that $P$ has degree at most one: $P(z) = az+b$. Then $\sin(-z) = -\sin(z)$ implies $e^{P(z)} = e^{P(-z)}$, i.e. $1 = e^{P(z)-P(-z)} = e^{2az}$, so that $a=0$  and $P$ is constant.

Then use $\frac{\sin(z)}{z} \to 1$ for $z\to0$ to conclude that $e^{P(z)} = 1$.