Proof of the summation $n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k$? $$n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k$$
Could anyone give the proof of the above equation? Thanks in advance!
 A: Note that, by commutativity, $$\sum_{k=0}^{n}\dbinom{n}{k}\left(n-k+1\right)^{n}\left(-1\right)^{k}=\sum_{k=0}^{n}\dbinom{n}{k}\left(k+1\right)^{n}\left(-1\right)^{n-k}
 $$ so let us consider $$\sum_{k=0}^{n}\dbinom{n}{k}x^{k+1}\left(-1\right)^{n-k}=x\left(x-1\right)^{n}
 $$ then if we differentiate and we multiply by $x$ we get $$\sum_{k=0}^{n}\dbinom{n}{k}\left(k+1\right)x^{k+1}\left(-1\right)^{n-k}=nx^{2}\left(x-1\right)^{n-1}+x\left(x-1\right)^{n}
 $$ and if we repeat the process $$\sum_{k=0}^{n}\dbinom{n}{k}\left(k+1\right)^{2}x^{k+1}\left(-1\right)^{n-k}=n\left(n-1\right)x^{3}\left(x-1\right)^{n-2}+2x^{2}n\left(x-1\right)^{n-1}+nx^{2}\left(x-1\right)^{n-1}+x\left(x-1\right)^{n}
 $$ and so if we repeat the process $n$ times we get $$\sum_{k=0}^{n}\dbinom{n}{k}\left(k+1\right)^{n}x^{k+1}\left(-1\right)^{n-k}=n!x^{n}+\textrm{addends with a positive power of }\left(x-1\right)
 $$ so if we evaluate the sum at $x=1$ we get 

$$\sum_{k=0}^{n}\dbinom{n}{k}\left(k+1\right)^{n}\left(-1\right)^{n-k}=\color{red}{n!}
 $$ 

as wanted.
A: Another variation of the theme. In the following we use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g.
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n\qquad\text{and}\qquad  k^n=n![z^n]e^{kz}
\end{align*}

We obtain
  \begin{align*}
\sum_{k=0}^n&\binom{n}{k}(n-k+1)^n(-1)^k\\
&=\sum_{k=0}^n\binom{n}{k}(k+1)^n(-1)^{n-k}\tag{1}\\
&=\sum_{k=0}^\infty[z^k](1+z)^nn![x^n]e^{(k+1)x}(-1)^{n-k}\tag{2}\\
&=(-1)^nn![x^n]e^x\sum_{k=0}^\infty(-e^x)^k[z^k](1+z)^n\tag{3}\\
&=(-1)^nn![x^n]e^x\left(1-e^x\right)^n\tag{4}\\
&=(-1)^nn![x^n]\left(-x\right)^n\tag{5}\\
&=n!\\
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we exchange the order of summation $k\rightarrow n-k$

*In (2) we extend the limit to infty without changing anything since we are adding zeros only. We apply the coefficient of operator to $\binom{n}{k}$ and to $(k+1)^n$ using $e^{(k+1)x}$.

*In (3) we do a simple rearrangement

*In (4) we use the substitution rule of the coefficient of operator
\begin{align*}
A(x)=\sum_{k=0}^\infty a_k x^k=\sum_{k=0}^\infty x^k [z^k]A(z)
\end{align*}

*In (5) we consider  the series expansion
\begin{align*}
e^x(1-e^x)^n&=(1+x+\cdots)\left(-x+\frac{x^2}{2}-\cdots\right)^n\\
&=\left((-x)^n\pm\text{powers of }x\text{ greater than }n\cdots\right)
\end{align*}
and need only to respect the $x$-term with power equal to $n$.
A: Suppose we seek to evaluate
$$\sum_{k=0}^n {n\choose k} (-1)^k (n-k+1)^n.$$
Introduce
$$(n-k+1)^n = 
\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\exp((n-k+1)z) \; dz.$$
We get for the sum
$$\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\exp((n+1)z) 
\sum_{k=0}^n {n\choose k} (-1)^k \exp(-kz)
\; dz
\\ = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\exp((n+1)z) 
(1-\exp(-z))^n
\; dz
\\ = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\exp(z) 
(\exp(z)-1)^n
\; dz.$$
This is $$n! [z^n] \exp(z) (\exp(z)-1)^n.$$
Now $$\exp(z)-1 = z + \frac{z^2}{2} + \frac{z^3}{6} +\cdots$$
and hence
$$(\exp(z)-1)^n = z^n + \cdots.$$
Therefore the result is
$$n! [z^0] \exp(z) = n!.$$
A: Let $A=\{1,2,\ldots, n\}$. The number of bijective functions $f:A\to A$ is clearly $n!$.
On the other hand, the number of functions from $A$ to a set with $n-k+1$ elements is $(n-k+1)^n$ and there are $\binom{n}{k}$ ways for choosing a subset of $A$ with $n-k$ elements.
So the given formula directly follows from the inclusion-exclusion principle.
Another chance is to use the forward difference operator $\delta$ bringing a polynomial $p(x)$ to $p(x+1)-p(x)$. If $p$ is not a constant we have that the degree of $\delta p$ is the degree of $p$ minus one, and the leading term of $\delta p$ equals the leading term of $p'$. It follows that $\delta^n$ applied to $p(x)=x^n$ gives the constant $n!$ for any $x$, with $(\delta^{n}p)(1)$ being the given sum.
