The complex function $z!$ is defined by
$z!=\int_{0}^{\infty}u^ze^{-u}du$
for Re($z$)$>-1$
For Re($z$)$≤ −1$ it is defined by
$z!=\frac{(z+n)!}{(z+n)(z+n-1)\cdot\cdot\cdot(z+1)}$
where $n$ is any (positive) integer $>$ −Re $z$. Being the ratio of two polynomials, $z!$ is analytic everywhere in the finite complex plane except at the poles that occur when $z$ is a negative integer.
Show that the definition of $z!$ for Re $z ≤ −1$ is independent of the value of $n$ chosen.
Proof:
Let $m$ and $n$ be two choices of integer with $m>n> −$Re $z$. Denote the corresponding definitions of $z!$ by $(z!)_m$ and $(z!)_n$ and consider the ratio of these two functions:
$\frac{(z!)_m}{(z!)_n}=\left(\frac{(z+m)!}{(z+m)(z+m-1)\cdot\cdot\cdot(z+1)}\right)\left(\frac{(z+n)(z+n-1)\cdot\cdot\cdot(z+1)}{(z+n)!}\right)=\frac{(z+m)!}{(z+m)(z+m-1)\cdot\cdot\cdot(z+n+1)\times(z+n)!}=\frac{(z+m)!}{(z+m)!}$
But I can't understand the second and third parts. I do not understand how that happens or what is used to get there. Could you help me?
The key point is that $(z+1)! = (z+1) z!$ for $\operatorname{Re} z > -1$ which can be shown using integration by parts. (This is equivalent to the functional equation $\Gamma(z+1) = z \Gamma(z)$ for $\operatorname{Re} z > 0$ of the Gamma function).
For $m > n > -\operatorname{Re} z$ we then have $$ (z+m)! = (z+m)(z+m-1)! = (z+m)(z+m-1)(z+m-2)! \\ = \ldots = (z+m)(z+m-1)\cdots(z+n+1) (z+n)! $$ and therefore $$ \require{cancel} (z!)_m = \frac{(z+m)!}{(z+m)\cdots (z+n+1)(z+n)\cdots (z+1)}\\ = \frac{\cancel{(z+m)}\cdots\cancel{(z+n+1)} (z+n)!}{\cancel{(z+m)}\cdots \cancel{(z+n+1)}(z+n)\cdots (z+1)} = (z!)_n $$ so that the definition is independent of the choice of $n$.