The infinite product

Let us start by noting the following two relations:

1.

For all $w_1,w_2\in D(0,K)$,

$$\vert\exp (w_1)-\exp (w_2)\vert \leq \exp (K)\vert w_1-w_2\vert$$ (1)

(this relation follows from the ML-formula and the fact that $\exp (z)^{\prime} =\exp (z)$).

2.

For $\vert z\vert\leq 1/2$,

$$\vert \Log (1+z) -z\vert \leq 2\vert z\vert^2$$

(this follows from the power series expansion of the principal logarithm and an estimation involving the geometric sum).

We shall now verify that

$$\prod_{n=1}^{\infty}(1+z/n)\exp (-z/n)$$

converges uniformly on compact subsets of the complex plane to a function P that has zeros exactly at the negative integers and that these zeros are simple.

For a given compact subset of the complex plane we choose N so large that the disk D(0,N) covers it. Then we work with this number N.

When $n\geq 2N$ and $\vert z\vert \leq N$ we have $\vert z/n\vert\leq 1/2$ and hence

$$\vert \Log (1+z/n) -z/n\vert \leq 2\vert z\vert^2/n^2.$$

Since $\sum 1/n^2 <\infty$ we know that there is a function, Q_N, analytic in D(0,N), such that

$$\sum_{n=2N}^M \left( \Log (1+z/n) -z/n\right) \to Q_N(z), \qquad \mbox{as } M\to \infty$$

uniformly on D(0,N).

Now we apply the exponential function and use () and find that

$$\prod_{n=2N}^M (1+z/n)\exp(-z/n) = \exp \left( \sum_{n=2N}^M \Log (1+z/n) -z/n \right) \to \exp (Q_N(z)),$$

as $M\to \infty$, uniformly on D(0,N). We notice that $\exp Q_N$ has no zeros in D(0,N).

Multiplication by the finite product

$$\prod_{n=1}^{2N-1}(1+z/n)\exp (-z/n)$$

does not affect the uniform convergence and we therefore obtain that

$$P(z)\equiv \prod_{n=1}^{\infty}(1+z/n)\exp (-z/n)$$

is analytic in D(0,N). It has simple zeros at $z=-1,-2,\cdots, -N+1$. The number N was chosen in order to cover a given compact set. Therefore we see that the infinite product defining P converges uniformly on compact subsets of the complex plane and hence that P is an entire function having its zeros at the negative integers.