The special case of Pochhammer Symbol at Zero? I am interested in a property of  Pochhammer Symbol. So I need an information about it.
Let  $a^{\bar{n}}$  Pochhammer symbol or rising factorial. As you know in the literature  $a^{\bar{0}}=1.$
I want to ask in a special case of $a=0$. For instance whether $0^{\bar{0}}$ or $0^{\bar{5}}$ are defined or not in literature.
In my opinion  $0^{\bar{0}}$=$1$ and $0^{\bar{5}}$=$0$ .
But when i have investigated about it, I am a little bit complicated becuase I have read in books of Rainville Special Functions in page 22. It is written as below:
For n is a natural number;
$\begin{aligned}(\alpha)^{\bar{n}} &=\prod_{k=1}^{n}(\alpha+k-1) \\ &=\alpha(\alpha+1)(\alpha+2) \cdots(\alpha+n-1), \quad n \geqq 1,  \\(\alpha)^{\bar{0}} &=1, \quad \alpha \neq 0 . \end{aligned}$
In another book of Mathai ve HauholdSpecial Functions for Applied scientists in page $1$:
$0^{\bar{5}}$ is not defined.
My open question; Can Pochhammer symbol or rising factorial be defined special case of zero?
My another question's link about this topic is  Rising factorials.
Thanks a lot for  your answers and references.
 A: (EDIT)
I would start with the 'extended' definition for any real value of $n$ :
$$\tag{1}x^{\bar{n}}=\frac{\Gamma(x+n)}{\Gamma(x)}$$
which should provide directly $x^{\bar{0}}=1$ (since $\Gamma(x)\not =0$ for $x\in \mathbb{R}$ and by analytic continuation for non positive integers as $\Gamma(x)$ becomes infinite)
Let's use $(1)$ too for $x^{\bar{n}}$ as $\,x\to 0$.
For $n>0$ the limit should give us $0^{\bar{n}}=0$ (because $\dfrac 1{\Gamma(x)}=x+\gamma x^2+O(x^3)\;$ to multiply by nearly $\Gamma(n)$)
For $n<0$ the results are more interesting :

*

*for $n$ not integer the results are $0$ again (as for $n>0$)

*for $n$ negative integer we get $\dfrac {(-1)^n}{(-n)!}$ (factorizing the numerator)

Should we consider the limit as $n\to 0$ then the result would be undefined since the limit would be between $1$ and $0$ depending of $x$ or $n$ going faster to $0$ (as $x\to 0$, $\dfrac 1{\Gamma(x)}\sim x$ so that forcing $n:=x$ would for example generate the limit $\dfrac 12$).
This written and if we set $n=0$ (instead of taking the limit as $n\to 0$) then we may apply our $x^{\bar{0}}=1$ result and obtain $0^{\bar{0}}=1$ (corresponding to our formula for negative integers) similar to the result for the ordinary power $0^{0}=1$ (and with the same reservations...).
A: Fix $n>0$ a natural number. Then by your definition $a^\bar{n}$ is a polynomial in $a$ with a root at $a=0$ coming from the first factor, and hence $0^\bar{n}=0$. (Note that the expression is not a polynomial if $n$ varies).
For the case $n=0$ by your definition we have $0^{\bar{0}} =\prod_{k=1}^0(k-1)$ and this empty product is equal to 1. However I would want to see how this matches up with other definitions before making a definitive statement here.
