Represent n! by binomial coefficients I find the following formula to be true
$$
\sum_{k=0}^n {{n}\choose{k}}(k+1)^n(-1)^k=(-1)^nn!
$$
at least for $n=1,2,3$, could any one give a proof about this in general? I'm almost sure it is true, because I use a different method to compare with a known result (in fact something of much higher level and in a very different field, the author said the result without proof, and I deduce that that result comes down to the above formula), for which I want to give an alternative proof. A good reference about this formula is preferred. I guess there should be a more general formula for which this is a special case. Thanks!
 A: If $f(x)\in R [x]$ is a polynomial over a unital ring $R$ and the degree of $f(x)$ is at most $n$, then
$$\sum_{k=0}^n\,(-1)^{n-k}\,\binom{n}{k}\,f(x+k)=n!\,a_n\,,$$
where $a_n$ is the coefficient of the $n$-th degree term of $f(x)$.  To show this, try to compute $\Delta^n\,f(x)$, where $\Delta$ is the forward differencing operator:
$$\Delta\,g(x):=g(x+1)-g(x)\,,$$
for $g(x)\in R[x]$.  The easiest way is probably induction on $n$, noting that (1) $\Delta$ decreases the degree and (2) $\Delta^n\,f(x)=\Delta^{n-1}\,\left(\Delta\,f(x)\right)$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\sum_{k = 0}^{n}{n \choose k}\pars{k + 1}^{n}\pars{-1}^{k} =
\sum_{k = 0}^{n}{n \choose k}\pars{-1}^{k}\
\overbrace{\bracks{n!\oint_{\verts{z} = 1}{\expo{\pars{k + 1}z} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}}}^{\ds{\pars{k + 1}^{n}}}
\\[5mm] = &\
n!\oint_{\verts{z} = 1}{\expo{z} \over z^{n + 1}}
\sum_{k = 0}^{n}{n \choose k}\pars{-\expo{z}}^{k}\,{\dd z \over 2\pi\ic} =
n!\oint_{\verts{z} = 1}{\expo{z} \over z^{n + 1}}
\pars{1 - \expo{z}}^{n}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\pars{-1}^{n}\,n!\oint_{\verts{z} = 1}{\expo{z} \over z^{n + 1}}\
\overbrace{\pars{\expo{z} - 1}^{n}}
^{\ds{n!\sum_{k = 0}^{\infty}{k \brace n}{z^{k} \over k!}}}\
\,{\dd z \over 2\pi\ic}\,,\qquad
\pars{\substack{\ds{i \brace j}\ \mbox{is a}
\\[1mm]
{\large Stirling\ Number\ of\ the\ Second\ Kind}}}
\\[5mm] = &\
\pars{-1}^{n}\,\pars{n!}^{2}\sum_{k = 0}^{\infty}\!\!{k \brace n}{1 \over k!}\
\overbrace{\oint_{\verts{z} = 1}{\expo{z} \over z^{n + 1 - k}}\,{\dd z \over 2\pi\ic}}^{\ds{1 \over \pars{n - k}!}} =
\pars{-1}^{n}\,\pars{n!}^{2}{n \brace n}{1 \over n!} = \bbx{\pars{-1}^{n}\,n!}
\end{align}

Reference: Stirling Number of the Second Kind.

A: Stirling Numbers of the Second Kind have the following relation:
$$
\newcommand{\stirtwo}[2]{\left\{#1\atop#2\right\}}
\begin{align}
(k+1)^n
&=\sum_{j=0}^n\stirtwo{n}{j}\binom{k+1}{j}\,j!\\
&=\sum_{j=0}^n\stirtwo{n}{j}\left[\binom{k}{j}+\binom{k}{j-1}\right]j!\\
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{k=0}^n\binom{n}{k}(k+1)^n(-1)^k
&=\sum_{k=0}^n(-1)^k\binom{n}{k}\sum_{j=0}^n\stirtwo{n}{j}\left[\binom{k}{j}+\binom{k}{j-1}\right]j!\\
&=\sum_{j=0}^n\stirtwo{n}{j}\sum_{k=0}^n(-1)^k\binom{n}{k}\left[\binom{k}{j}+\binom{k}{j-1}\right]j!\\
&=\sum_{j=0}^n\stirtwo{n}{j}\sum_{k=0}^n(-1)^k\left[\binom{n}{j}\binom{n-j}{k-j}+\binom{n}{j-1}\binom{n-j+1}{k-j+1}\right]j!\\
&=\sum_{j=0}^n\stirtwo{n}{j}(-1)^n\big([j=n]+[j=n+1]\big)\,j!\\[9pt]
&=(-1)^nn!
\end{align}
$$
since $\stirtwo{n}{n}=1$, and for $j\gt n$, $\stirtwo{n}{j}=0$.
