Show that $\sum_{k=0}^{n}(-1)^{n+k}{n\choose k}(ak+b)_n=n!a^n$ How would we go about showing that?

$$\sum_{k=0}^{n}(-1)^{n+k}{n\choose k}(ak+b)_n=n!a^n\tag1$$

$(x)_n=x(x+1)(x+2)\cdots(x+n-1)$
$(x)_0=1$
Recall from Binomial theorem 
$$(x+a)^n=\sum_{k=0}^{n}{n\choose k}x^ka^{n-k}\tag2$$
Setting $a=-1$
$$(x-1)^n=\sum_{k=0}^{n}(-1)^{n-k}{n\choose k}x^k\tag3$$
Maybe differentiates $(3)$ by m times?
$$n(x-1)^{n-1}=\sum_{k=0}^{n}(-1)^{n-k}{n\choose k}kx^{k-1}\tag4$$
$$n(n-1)(x-1)^{n-2}=\sum_{k=0}^{n}(-1)^{n-k}{n\choose k}k(k-1)x^{k-2}\tag5$$
an so on...but I can't see how to get to $(1)$
 A: Let we define the following (difference) operator:
$$\delta : p(x) \mapsto (\delta p)(x)=p(x+1)-p(x).$$
There are some simple properties if $p(x)$ is a polynomial with degree $d\geq 1$:


*

*The degree of $(\delta p)(x)$ is $d-1$;

*If the leading term of $p(x)$ is $ax^d$, the leading term of $(\delta p)(x)$ is the same as the leading term of $p'(x)$, namely $ad x^{d-1}$.


What happens if $p(x)$ is a polynomial with degree $n$ and we apply $\delta^n$ to it? By 1. we get a constant polynomial, and by 2. such constant polynomial is $n!$ times the leading coefficient of $p(x)$. So if we consider $p(x)=(ax+b)_n$ we get that $(\delta^n p)(x)=n!a^n$. On the other hand:
$$ (\delta^n p)(x) = \sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k} p(x+k) $$
is simple to prove by induction, and
$$ \sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k} p(k) = (\delta^n p)(0) = n! a^n $$
proves the claim.
A: We re-write this as
$$S_{a,b}(n) = \sum_{k=0}^n (-1)^{n+k} {n\choose k}
(ak+b+n-1)^{\underline{n}} = n! \times a^n.$$
This yields for the LHS
$$\sum_{k=0}^n (-1)^{n+k} {n\choose k}
\sum_{q=0}^n (-1)^{n+q} \left[n\atop q\right] (ak+b+n-1)^q.$$
Now clearly the original LHS is  a polynomial of degree $n$ in $a$. We
have the result  if we manage to evaluate  the coefficients on $[a^p]$
where $0\le p\le n$ which are given by
$$\sum_{k=0}^n (-1)^{n+k} {n\choose k}
\sum_{q=p}^n (-1)^{n+q} \left[n\atop q\right]
{q\choose p} k^p (b+n-1)^{q-p}.$$
This is
$$\sum_{q=p}^n (-1)^{q} \left[n\atop q\right]
{q\choose p} (b+n-1)^{q-p}
\sum_{k=0}^n (-1)^{k} {n\choose k} k^p.$$
With $k^p = p! [z^p] \exp(kz)$ we have
$$\sum_{k=0}^n (-1)^{k} {n\choose k} k^p
= p! [z^p] \sum_{k=0}^n (-1)^{k} {n\choose k} \exp(kz)
\\ = p! [z^p] (1-\exp(z))^n$$
and get for the sum
$$p! [z^p] \sum_{q=p}^n (-1)^{q} \left[n\atop q\right]
{q\choose p} (b+n-1)^{q-p} (1-\exp(z))^n
\\ = p! [z^p]  (1-\exp(z))^n
\sum_{q=p}^n (-1)^{q} \left[n\atop q\right]
{q\choose p} (b+n-1)^{q-p}.$$
Observe  that   $1-\exp(z)  =   -z  -  \frac{1}{2}   z^2  -\frac{1}{6}
z^3-\cdots$ hence we get zero for $p\lt n.$ For $p=n$ we are left with
$$n! (-1)^n \sum_{q=n}^n (-1)^{q} \left[n\atop q\right]
{q\choose n} (b+n-1)^{q-n}
\\ = n! \left[n\atop n\right] {n\choose n} \times 1
= n!$$
This means that
$$[a^p] S_{a,b}(n) = [[p = n]] \times n!$$
and we have the claim.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\sum_{k = 0}^{n}\pars{-1}^{n + k}{n \choose k}\pars{ak + b}_{n} =
\pars{-1}^{n}\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\,
{\Gamma\pars{ak + b + n} \over \Gamma\pars{ak + b}}
\\[5mm] = &\
\pars{-1}^{n}\,n!\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}{ak + b + n - 1 \choose n} =
\pars{-1}^{n}\,n!\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}
\bracks{z^{n}}\pars{1 + z}^{ak + b + n - 1}
\\[5mm] = &\
\pars{-1}^{n}\,n!\bracks{z^{n}}\braces{\pars{1 + z}^{b + n - 1}
\sum_{k = 0}^{n}{n \choose k}\bracks{-\pars{1 + z}^{a}}^{k}}
\\[5mm] = &\
\pars{-1}^{n}n!\bracks{z^{n}}\braces{\pars{1 + z}^{b + n - 1}
\bracks{1 - \pars{1 + z}^{a}}^{n}} =
n!\bracks{z^{0}}\braces{\pars{1 + z}^{b + n - 1}
\bracks{\pars{1 + z}^{a} - 1 \over z}^{n}}
\\[5mm] = &\
n!\,\lim_{z \to 0}\braces{\pars{1 + z}^{b + n - 1}
\bracks{\pars{1 + z}^{a} - 1 \over z}^{n}} = \bbx{\ds{n!\,a^{n}}}
\end{align}

because $\ds{\lim_{z \to 0}{\pars{1 + z}^{a} - 1 \over z} = \color{#f00}{a}}$.

A: Note that $$S\left(a,b\right)=\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\left(ak+b\right)_{n}=a^{n}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k+n}\left(k+\frac{b}{a}\right)_{n}$$ $$=a^{n}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\left(-k-\frac{b}{a}\right)_{n}=-a^{n}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{\left(-k-\frac{b}{a}\right)\left(-k-\frac{b}{a}\right)_{n}}{k+\frac{b}{a}}$$ so setting $$y=-\frac{b}{a},\,x=\frac{b}{a},\,f\left(z\right)=z\left(z\right)_{n}$$ from the Melzak's identity for polynomial of degree $n+1$ $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{f\left(y-k\right)}{x+k}=\frac{f\left(x+y\right)}{x\dbinom{x+n}{n}}-n!a_{n+1},\,x\neq-k$$ where $a_{n+1}$ is the coefficient of $z^{n+1}$, we get $$S\left(a,b\right)=-a^{n}\left(\frac{f\left(0\right)}{\frac{b}{a}\dbinom{\frac{b}{a}+n}{n}}-n!\right)=\color{red}{a^{n}n!}$$ as wanted.
