Difference Calculus of the Factorial Function for Negative Integer Powers of $x$ Under the difference calculus, we have that 
$$\Delta f(x)=f(x+1)-f(x)$$.
We also have factorial polynomials (sometimes referred to the rising and or falling factorials).  The text I am reading defines the factorial function as
$$x^{(n)}=x(x-1)(x-2)...(x-(n-1))$$
with $x^{(0)}=1$ by definition.  Applying Delta to the factorial function gives
$$\Delta x^{(n)}=(x+1)^{(n)}-x^{(n)}$$
$$=(x+1)[x(x-1)...(x-(n-2))-[x(x-1)...(x-(n-2))](x-(n-1))$$
$$=[(x+1)-(x-n+1)][x(x-1)...(x-(n-2))]$$
$$=nx^{(n-1)}$$
and this holds for positive $n$.  It is then noted that 
$$x^{(m+n)}=x(x-1)...(x-(m-1))\cdot(x-m)(x-m-1)...(x-m-(n-1))$$
and so
$$x^{(m+n)}=x^{(m)}\cdot (x-m)^{(n)}$$
Now by the above, if we let $m+n=0$, then $m=-n$. So we have
$$1=x^{(0)}=x^{(-n+n)}=x^{(-n)}(x-(-n))^{(n)}=x^{(-n)}(x+n)^{(n)}$$ 
So since $1=x^{(-n)}(x+n)^{(n)}$, we have that 
$$x^{(-n)}=\frac{1}{(x+n)^{(n)}}$$
Now I am in the game to show that
$$\Delta x^{(n)}=nx^{(n-1)}$$
for negative integers.  Which I interpret as trying to find 
$$\Delta x^{(-n)}$$  So when I attempt the calculation, I get 
$$\Delta x^{(-n)}=(x+1)^{(-n)}-x^{(n)}$$
$$=\frac{1}{(x+1+n)^{(n)}}-\frac{1}{(x+n)^{(n)}}$$
$$=\frac{1}{(x+1+n)(x+n)...(x+2)}-\frac{1}{(x+n)(x+(n-1))...(x+1)}$$
$$=\frac{1}{(x+1+n)[(x+n)...(x+2)]}-\frac{1}{[(x+n)...(x+2)](x+1)}$$
$$=\frac{1}{x+1+n}\frac{1}{(x+n)...(x+2)}-\frac{1}{x+1}\frac{1}{(x+n)...(x+2)}$$
$$=\left[\frac{1}{x+1+n}-\frac{1}{x+1}\right]\frac{1}{(x+n)...(x+2)}$$
$$=\frac{(x+1)-(x+1+n)}{(x+1)(x+1+n)}\frac{1}{(x+n)^{(n-1)}}$$
And this is where i get stuck...  


Everything looks fine: We obtain
  \begin{align*}
\color{blue}{\Delta x^{(-n)}}&=(x+1)^{(-n)}-x^{(n)}\\
&=\cdots\\
&=\frac{(x+1)-(x+1+n)}{(x+1)(x+1+n)}\cdot\frac{1}{(x+n)^{(n-1)}}\\
&=-n\cdot\frac{1}{(x+n+1)\cdot (x+n)^{(n-1)}\cdot (x+1)}\\
&=-n\cdot\frac{1}{(x+n+1)^{(n+1)}}\\
&=\color{blue}{-nx^{(-n-1)}}
\end{align*}
  and we conclude the rule is valid for negative $n$ as well.

Hint: You might be interested in the calculation following (2.52) in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.
