Q:

Hadamard product for $e^z - 1$

The zeros are $\{2 \pi i n ,n\in \mathbb{Z}\}$. They are all simple. The function is of order 1. Hence, the factorization should be 
$$ e^z - 1 = e^{p(z)} z \prod_{n=1}^\infty (1 + z^2/4n^2\pi^2) . $$
The problem left is to determine the polynomial $p(z)$, which is at most first order. 
But how to proceed? 

A:

Let $p(z)=az+b$.
$\frac{e^z-1}{z}$ goes to $1$ at zero so $b=0$ by taking that limit on RHS after dividing by $z$.
Then the coefficient of $z$ on the left is $\frac{1}{2}$ while the product on the right starts at $z^2$ so we need $a=\frac{1}{2}$

