TITLE: how to prove this combinatorial identity I accidentally find?
QUESTION [7 upvotes]: Today when I solve a counting problem using different methods I find the following (seemingly correct) combinatorial identity, but I can't find it on the Internet and I can't prove its correctness neither. But I have verified its correctness with positive integer $n$ within $[0, 1000]$ using a simple computer program. Anyone can give a proof to this identity (or any link to its proof)?
$$\frac{(n+1)n}{2} \cdot n! = \sum\limits_{k=0}^{n} (-1)^k \cdot \frac{n!}{k!\cdot(n-k)!} \cdot (n-k)^{n+1}$$
And equivalently if you want,
$$\frac{(n+1)n}{2} = \sum\limits_{k=0}^{n} (-1)^k \cdot \frac{1}{k!\cdot(n-k)!} \cdot (n-k)^{n+1}$$

REPLY [4 votes]: Here   is another variation of the theme. It   is   convenient to use  the  coefficient  of operator $[t^k]$ to denote the coefficient of $t^k$ in a series. This way we can write e.g.
\begin{align*}
\binom{n}{k}=[t^k](1+t)^n\qquad\text{and}\qquad  k^n=n![t^n]e^{kt}
\end{align*}

OPs identity can be written as
\begin{align*}
\frac{n}{2}(n+1)!=\sum_{k=0}^n&(-1)^k\binom{n}{k}(n-k)^{n+1}\quad\qquad n\geq 0
\end{align*}
We      start with the right-hand side and obtain
  \begin{align*}
\sum_{k=0}^n&(-1)^k\binom{n}{k}(n-k)^{n+1}\\
&=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^{n+1}\tag{1}\\
&=\sum_{k=0}^\infty(-1)^{n-k}[u^k](1+u)^n(n+1)![t^{n+1}]e^{kt}\tag{2}\\
&=(-1)^n(n+1)![t^{n+1}]\sum_{k=0}^\infty(-e^t)^k[u^k](1+u)^n\tag{3}\\
&=(-1)^n(n+1)![t^{n+1}]\left(1-e^t\right)^n\tag{4}\\
&=(n+1)![t^{n+1}]\left(t+\frac{t^2}{2}+\cdots\right)^n\tag{5}\\
&=\frac{n}{2}(n+1)!\tag{6}
\end{align*}
  and the claim follows.

Comment:

In (1) we change the order of  summation: $k\longrightarrow n-k$
In (2) we apply the coefficient of operator twice. We also extend the range of summation to infinity without changing anything, since we are adding zeros only.
In (3) we do some rearrangements and use the linearity of the coefficient of operator.
In (4) we use the substitution rule with $u=-e^t$
\begin{align*}
A(t)=\sum_{k=0}^\infty a_kt^k=\sum_{k=0}^\infty t^k[u^k]A(u)\\
\end{align*}
In (5) we factor out $(-1)^n$ and expand the exponential series.
In (6) we note that in order to extract the coefficient of $t^{n+1}$ we have $n$ possibilities to select the term $t$ and one possibility to select the the term $\frac{t^2}{2}$ giving $\frac{n}{2}$.