(related to Definition: Falling And Rising Factorial Powers)
Now, we want to motivate, why we have provided the given formulae for negative factorial powers. The reason for setting $$x^{\underline 0}:=1,\quad x^{\underline{-n}}:=\frac{1}{(x+n)^{\underline{n}}}$$ and $$x^{\overline 0}:=1,\quad x^{\overline{-n}}:=\frac{1}{(x-n)^{\overline{n}}}$$ is similar to the reason given in elementary algebra that, for given positive integers $n$ and $m$ we have the formula $x^nx^m=x^{n+m},$ and the only way to keep this formula hold also for negative exponents is to set $x^0=1$ and $x^{-n}=1/x^n.$
Note that by definition of falling factorial powers for positive $n,m\ge 1$ we have $$\begin{align}x^{\underline{n}}(x-n)^{\underline{m}}&=x(x-1)(x-2)\cdots(x-n+2)(x-n+1)\cdot \nonumber \\ &\quad\cdot (x-n)(x-n-1)(x-n-2)\ldots(x-n-m+2)(x-n-m+1) \nonumber\\ &=x^{\underline{n+m}}.\nonumber\end{align}$$
If this formula has to hold also for $n=0$, then we have to require $x^{\underline 0}=1,$ because then $$x^{\underline 0}(x-0)^{\underline m}=x^{\underline {0+m}}=x^{\underline 0}\cdot x^{\underline m}=1\cdot x^{\underline m}.$$
Therefore, for every integer $n\in Z$ we require $$x^{\underline {-n}}(x-(-n))^{\underline {n}}=x^{\underline {-n+n}}=x^{\underline {0}}=1,$$ and thus $$x^{\underline{-n}}:=\frac{1}{(x+n)^{\underline{n}}}.$$
Analogously, by definition of raising factorial powers for positive $n,m\ge 1$ we have $$\begin{align}x^{\overline{n}}(x-n)^{\overline{m}}&=x(x+1)(x+2)\cdots(x+n-2)(x+n-1)\cdot \nonumber \\ &\quad\cdot (x+n)(x+n+1)(x+n+2)\ldots(x+n+m-2)(x+n+m-1) \nonumber\\ &=x^{\overline{n+m}}.\nonumber\end{align}$$
If this formula has to hold also for $n=0$, then we have to require $x^{\overline 0}=1,$ because then $$x^{\overline 0}(x+0)^{\overline m}=x^{\overline{0+m}}=x^{\overline 0}\cdot x^{\overline m}=1\cdot x^{\overline m}.$$
Thus, for every integer $n\in Z$ we require $$x^{\overline{-n}}(x+(-n))^{\overline {n}}=x^{\overline{-n+n}}=x^{\overline{0}}=1,$$ and thus $$x^{\overline{-n}}:=\frac{1}{(x-n)^{\overline{n}}}.$$
