Prove that $n! = \sum_{r=0}^{n}\left((-1)^{r}\left({}^{{n}}\mathrm{C}_{r}\right)\left(n-r\right)^{n}\right)$ 
Prove that $n!=\sum_{r=0}^{n}\left((-1)^{r}\left({}^{{n}}\mathrm{C}_{r}\right)\left(n-r\right)^{n}\right)$

I have a method to solve this and here it is :
Number of onto functions from $A\rightarrow B$ where $n(A) = m , n(B) =n, $ where $m\ge n$ is given by

$f(m,n)=n^m -{}^{{n}}\mathrm{C}_{1}\left(n-1\right)^{m}+{}^{{n}}\mathrm{C}_{2}\left(n-2\right)^{m}-\cdots$

so now if $n(A)=n(B)=n$ we get number of onto functions $= f(n,n)$
But note that if $n(A)=n(B)=n$ then all the onto functions are also one - one functions and all one -one functions are onto.
Now number of one-one function $=n!$
$\implies n! =f(n,n)=n^n -{}^{{n}}\mathrm{C}_{1}\left(n-1\right)^{n}+{}^{{n}}\mathrm{C}_{2}\left(n-2\right)^{n}-\cdots=\sum_{r=0}^{n}\left((-1)^{r}\left({}^{{n}}\mathrm{C}_{r}\right)\left(n-r\right)^{n}\right)$
$\boxed{\therefore n! =\sum_{r=0}^{n}\left((-1)^{r}\left({}^{{n}}\mathrm{C}_{r}\right)\left(n-r\right)^{n}\right)}$
But I want some another method to solve this, without using the number of functions logic.
Thanks for any help in advance !
 A: Let $ n\in\mathbb{N} $, and $ f_{n} : x\mapsto\left(\operatorname{e}^{x}-1\right)^{n} $, let's try to expand $ f_{n} $ to the $ n^{\text{th}} $ order, around $ 0 $ :
On the one hand, we have : $$ f_{n}\left(x\right)=\left(x+\underset{x\to 0}{\overset{\mathcal{O}}{}}\left(x\right)\right)^{n}=x^{n}+\underset{x\to 0}{\overset{\mathcal{O}}{}}\left(x^{n}\right) $$
On the other we have : \begin{aligned} f_{n}\left(x\right)=\sum_{k=0}^{n}{\left(-1\right)^{n-k}\binom{n}{k}\operatorname{e}^{kx}}&=\sum_{k=0}^{n}{\left(-1\right)^{n-k}\binom{n}{k}\sum_{i=0}^{n}{\frac{k^{i}}{i!}x^{i}}}+\underset{x\to 0}{\overset{\mathcal{O}}{}}\left(x^{n}\right) \\ &=\sum_{i=0}^{n}{\frac{1}{i!}\left(\sum_{k=0}^{n}{\left(-1\right)^{n-k}\binom{n}{k}k^{i}}\right)x^{i}}+\underset{x\to 0}{\overset{\mathcal{O}}{}}\left(x^{n}\right)\end{aligned}
Since, the expansion must be unique, $ \left(\forall i<n\right),\ \frac{1}{i!}\sum\limits_{k=0}^{n}{\left(-1\right)^{n-k}\binom{n}{k}k^{i}}=0$, and for $i=n $, we have : $$ \fbox{$\begin{array}{rcl}\displaystyle\sum_{k=0}^{n}{\left(-1\right)^{n-k}\binom{n}{k}k^{n}}=n!\end{array}$} $$
Another way to prove it using the same function, is to prove, first of all, by induction, that : $$ \left(\forall p\leq n\right),\ f_{n}^{\left(p\right)}\left(x\right)=\sum_{k=1}^{p}{a_{p,k}\frac{n!}{\left(n-k\right)!}\operatorname{e}^{kx}\left(\operatorname{e}^{x}-1\right)^{n-k}} $$
Where $ \left(a_{p,k}\right)_{0\leq k\leq p\leq n} $ are such that : $$ a_{p,1}=1, a_{p,p}=1, \left(\forall k\in\left[\!\left[2,p\right]\!\right]\right),\ a_{p+1,k}=ka_{p,k}+a_{p,k-1} $$
If we took a look at the case $ p=n $, we can see that $ f_{n}^{\left(n\right)}\left(0\right)=n! $.
If we expand $ f_{n} $ using the binomial theorem, then looked at $ f_{n}^{\left(n\right)}\left(0\right) $, we get that $ f_{n}^{\left(n\right)}\left(0\right)=\sum\limits_{k=0}^{n}{\left(-1\right)^{n-k}\binom{n}{k}k^{n}} $. Hence the equality.
