"Chain Rule" for Finite Difference Operator I am trying to prove using difference calculus the following formula;
$$\Delta(a+bx)^{(n)}=bn(a+bx)^{(n-1)}$$
which is akin to the chain rule for continuous Calculus.  Here we have that 
$$x^{(n)}=x(x-1)...(x-n+1)$$
and
$$\Delta f(x)=f(x+1)-f(x)$$
It seemed daunting at first so I tried small values; $a=3, b=2$.  Looking at the relationship with actual values would help generate an idea how the general proof would look.  So I started;
$$\begin{eqnarray*}\Delta(3+2x)^{(n)}&=&[3+2(x+1)]^{(n)}-(3+2x)^{(n)}\\&=&(5+2x)^{(n)}-(3+2x)^{(n)}\\&=&(2x+5)(2x+4)(2x+3)^{(n-2)}-(2x+3)^{(n-2)}(2x+5-n)(2x+4-n)\\&=&(2x+3)^{(n-2)}[(2x+5)(2x+4)-(2x+5-n)(2x+4-n)]\\&=&(2x+3)^{(n-2)}[n(2x+5)+n(2x+4)-n^2]\\&=&n(2x+3)^{(n-2)}[(2x+5)+(2x+4)-n]\\&=&n(2x+3)^{(n-2)}(4x+9-n)
\end{eqnarray*}$$
This last couple of steps is escaping me.  I can't really factor a 2 without ending up with a $1/2$n.  I know that if I can manipulate the last term into something close to $2(2x+3-(n-2))$ then this should give me what I need, but this expanded is $$(4x+6-2n+4)=4x+10-2n=4x+9-n+(1-n)$$ and so I'm missing a $(1-n)$.
DId I screw up some algebra?  
EDIT;  @ancientmathematician, you were right, and my definition of $p(x)^{(n)}$ was incorrect, and is defined as
$$p(x)^{(n)}=p(x)p(x-1)...p(x-n+1)$$
which I'm sure will change things around.  I will go through the algebra shortly and If I solve it i will answer my own question, or someone else can write it up and I'll upvote it.

As you mentioned, the algebra works out when using the definition you gave in the edit. 
Let $h(x) = f(x)^{(n)} = f(x)f(x-1)\cdots f(x-n+1)$. Then
\begin{align}
\Delta h(x) &= f(x+1) f(x) f(x-1) \cdots f(x-n+2) - f(x)f(x-1)\cdots f(x-n+1) \\
&= f(x) f(x-1) \cdots f(x-n+2)(f(x+1) - f(x-n+1)) \\
&= (f(x+1) - f(x-n+1)) f(x)^{(n-1)}.
\end{align}
In the special case that $f(x) = a + bx$, we have
\begin{align}
f(x+1) - f(x-n+1) &= a + bx + b - (a + bx - (n-1) b) \\
&= nb.
\end{align}
So we find that
$$
\Delta h(x) = nb (a + bx)^{(n-1)}.
$$
