Evaluating a Difference Operator of a Product of Factorial Polynomials I'm looking for a verification on a Difference Operator question.  

Evaluate the following difference;  $$\Delta x^{(4)}x^{(-3)}$$

I actually did it a couple of different ways, but as there is no answer key, I would just like to know that it is correct.  For the problem, we need that 
$$x^{(n)}=x(x-1)...(x-n+1)$$
$$x^{(-n)}=\frac{1}{(x+n)^{(n)}}$$
and 
$$\Delta{f(x)}=f(x+1)-f(x)$$
Now, for the solution;
$$\begin{eqnarray*}\Delta x^{(4)}x^{(-3)}&=&(x+1)^{(4)}(x+1)^{(-3)}-x^{(4)}x^{(-3)}\\&=&\frac{(x+1)^{(4)}}{(x+4)^{(3)}}-\frac{x^{(4)}}{(x+3)^{(3)}}\\&=&\frac{(x+1)^{(4)}(x+1)-x^{(4)}(x+4)}{(x+4)^{(4)}}\\&=&\frac{x^{(3)}(x+1)(x+1)-x^{(3)}(x+4)(x-3)}{(x+4)^{(4)}}&=&\frac{x^{(3)}[(x^2+2x+1)-(x^2+x-12)]}{(x+4)^{(4)}}\\&=&x^{(3)}x^{(-4)}(x+13)
\end{eqnarray*}$$
While I got the same answer using a Difference operator product identity
$$\Delta[f(x)g(x)]=g(x)\Delta f(x)+f(x+1)\Delta g(x)$$
the final factor seems out of place, so I just want to verify its existence.
 A: Everything looks fine.

We obtain on the one hand
  \begin{align*}
\color{blue}{\Delta\left(x^{(m)}x^{(-n)}\right)}&=(x+1)^{(m)}(x+1)^{(-n)}-x^{(m)}x^{(-n)}\\
&=\frac{(x+1)^{(m)}}{(x+n+1)^{(n)}}-\frac{x^{(m)}}{(x+n)^{(n)}}\\
&=\frac{(x+1)x^{(m-1)}(x+1)}{(x+n+1)^{(n+1)}}-\frac{x^{(m-1)}(x-m+1)(x+n+1)}{(x+n+1)^{(n+1)}}\\
&=x^{(m-1)}x^{(-n-1)}\left[(x+1)^2-(x+1-m)(x+1+n)\right]\\
&\,\,\color{blue}{=x^{(m-1)}x^{(-n-1)}\left[(m-n)x+mn+m-n\right]}
\end{align*}

$$ $$

On the other hand we obtain
\begin{align*}
\color{blue}{x^{(-n)}}&\color{blue}{\Delta x^{(m)}+(x+1)^{(m)}\Delta x^{(-n)}}\\
&=x^{(-n)}\left[(x+1)^{(m)}-x^{(m)}\right]+(x+1)^{(m)}\left[(x+1)^{(-n)}-x^{(-n)}\right]\\
&=\frac{(x+1)^{(m)}-x^{(m)}}{(x+n)^{(n)}}+(x+1)^{(m)}\left[\frac{1}{(x+n+1)^{(n)}}-\frac{1}{(x+n)^{(n)}}\right]\\
&=\frac{x^{(m-1)}}{(x+n+1)^{(n+1)}}\left[(x+1)-(x-m+1)\right](x+n+1)\\
&\qquad+\frac{x^{(m-1)}}{(x+n+1)^{(n+1)}}(x+1)\left[(x+1)-(x+n+1)\right]\\
&=x^{(m-1)}x^{(-n-1)}m\left(x+n+1\right)-x^{(m-1)}x^{(-n-1)}n(x+1)\\
&\,\,\color{blue}{=x^{(m-1)}x^{(-n-1)}\left[(m-n)x+mn+m-n\right]}
\end{align*}

Both equality chains coincide and putting $m=4$ and $n=3$ we finally get
\begin{align*}
\Delta\left(x^{(4)}x^{(-3)}\right)=x^{(3)}x^{(-4)}\left(x+13\right)
\end{align*}
in accordance with OP's result.
