“This post is dedicated to my beloved little brother~~, who's getting wrecked in the military~~.”
A factorial is defined as follows:
$$ z! = \begin{cases} z \times (z-1) \times (z-2) \times \dots \times 2 \times 1 &(z \in \mathbb{N}) \\ 1 &(z=0) \end{cases}=\prod^z_{n=1}{n}\quad(z\in\mathbb{N_0}) $$
However, since this definition includes a product notation whose upper index is $z$ (i.e., $\displaystyle\prod^z$) , it is not well-defined when $z$ is not a nonnegative integer, thereby obstructing any extension of the definition beyond non-negative integers.
In this paper, we derive a new definition—equivalent to the conventional factorial but without the $\displaystyle\prod^z$ notation—enabling us to extend the factorial to the complex domain.
The factorial satisfies the following recurrence relation:
$$ \begin{align} (z+1)!=& \, z!\times (z+1)\quad(z\in\mathbb{N_0}) \end{align} $$
Thus, the factorial of “the next integer after $z$” is defined by multiplying $z!$ by that next integer.
Extending this definition repeatedly “to the next integer” yields a generalized form for the factorial of “the $N$-th next integer after $z$”:
$$ \begin{align} (z+1)!=& z!\times (z+1)\\ (z+2)!=& z!\times (z+1)(z+2)\\ &\vdots\\ (z+N)!=& z!\times (z+1)(z+2)\cdots(z+N)\\ \therefore(z+N)!=&z!\prod^N_{n=1}{(z+n)}\quad(N\in\mathbb{N_0}) \end{align} $$
Using the fact that “the $N$-th next integer after $z$” is the same as “the $z$-th next integer after $N$,” we have:
$$ \begin{align} (z+N)!= z!\prod^N_{n=1}{(z+n)}\;\Longleftrightarrow\; (N+z)!= N!\prod^z_{n=1}{(N+n)} \end{align} $$
For nonnegative integers $z$ and $N$, this yields the following identity:
$$ \begin{align} \therefore z!\prod^N_{n=1}{(z+n)}=N!\prod^z_{n=1}{(N+n)}\quad(z,\,N\in\mathbb{N_0}) \end{align} $$
Rearranging so that only $z!$ remains on the left and simplifying appropriately, we obtain:
$$ \begin{align} z!&=\frac{N!\displaystyle\prod^z_{n=1}{(N+n)}}{\displaystyle\prod^N_{n=1}{(z+n)}}\\ &=\frac{\left(\displaystyle\prod^N_{n=1}{n}\right)\left(\displaystyle\prod^z_{n=1}{(N+n)}\right)}{\displaystyle\prod^N_{n=1}{(z+n)}}\\ &=\left(\displaystyle\prod^N_{n=1}{\frac{n}{z+n}}\right)\left(\displaystyle\prod^z_{n=1}{(N+n)}\right)\\ &=\left(\prod^N_{n=1}{\frac{1}{1+\frac{z}{n}}}\right)\left(\displaystyle\prod^z_{n=1}{(N+1)\cdot\frac{N+n}{N+1}}\right)\\ &=\left(\prod^N_{n=1}{\frac{1}{1+\frac{z}{n}}}\right)\left(\displaystyle\prod^z_{n=1}{(N+1)}\right)\left(\displaystyle\prod^z_{n=1}{\frac{N+n}{N+1}}\right)\\ &=\left(\prod^N_{n=1}{\frac{1}{1+\frac{z}{n}}}\right)(N+1)^z\left(\displaystyle\prod^z_{n=1}{\frac{N+n}{N+1}}\right) \end{align} $$