<s>[INST] "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.
 [/INST] 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
$$
\Delta h(x) = nb (a + bx)^{(n-1)}.
$$
</s>