The $m$th falling factorial power of $x$ is defined as
$x^{\underline m}:=x(x-1)...(x-m+1),$
and the difference operator as
$\Delta f(x) := f(x+1)-f(x).$
One fundamental statement in finite calculus is the identity
$\Delta x^{\underline m} = m x^{\underline {m-1}}.$
However, there is a more general version (to be found, e.g. here, page 68, eq.(6)): Let $a,b\in\mathbb{Z}$, then
$\Delta (ax+b)^{\underline m}=am(ax+b)^{\underline {m-1}}.$
Unfortunately (to my understanding) the author gives no clear justification. Is there a simple way to derive this from the previously mentioned $(a,b)=(1,0)$-case. The shift $b$ seems to be trivial, but I have no idea how to understand $a\neq 1$.
Edit:
As @BrianM.Scott pointed out correctly, the statement is false. As a final remark I would like to provide the correct version of the statement, which is true, to solve the misunderstanding of the Book cited above.
Define the more general $m$th falling factorial power with step $s$ as
$x^{\underline {m,s}}:=x(x-s)(x-2s)...(x-(m-1)s),$
then
$\Delta (ax+b)^{\underline {m,a}} = am(ax+b)^{\underline {m-1,a}}.$
This can be easily proven by induction in $m$.
It’s not true in general:
$$\begin{align*} \Delta(2x)^{\underline2}&=(2(x+1))^{\underline2}-(2x)^{\underline2}\\ &=(2x+2)(2x+1)-2x(2x-1)\\ &=(4x^2+6x+2)-(4x^2-2x)\\ &=8x+2\\ &\ne 8x\\ &=2\cdot2(2x)^{\underline 1} \end{align*}$$