Proving Combinatorical Summation: $n!=\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n$ been stuck with this question for the last few hours, any help would be appreciated.
$$
{\large n! = \sum_{k = 0}^{n}\left(-1\right)^{k}{\,n\, \choose \,k\,}
\left(\,n - k\,\right)^{n}}
$$
what I did:
$\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n=n!\sum_{k=0}^n\frac{(-1)^k(n-k)^n}{k!(n-k)!}.$
So we are left to prove $\sum_{k=0}^n\frac{(-1)^k(n-k)^n}{k!(n-k)!}=1$. tried doing so using induction, or treating the sum as geometric sequence (which turned out poorly)
Suggestions?
 A: For $j\lt n$,
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{m+k}{j}
&=\sum_{k=0}^n(-1)^k\binom{n}{k}\sum_{i=0}^j\binom{m}{j-i}\binom{k}{i}\tag{1}\\
&=\sum_{i=0}^j\binom{m}{j-i}\sum_{k=0}^n(-1)^k\binom{n}{k}\binom{k}{i}\tag{2}\\
&=\sum_{i=0}^j\binom{m}{j-i}\sum_{k=0}^n(-1)^k\binom{n}{i}\binom{n-i}{k-i}\tag{3}\\
&=\sum_{i=0}^j\binom{m}{j-i}\binom{n}{i}(-1)^i\sum_{k=0}^n(-1)^{k-i}\binom{n-i}{k-i}\tag{4}\\
&=\sum_{i=0}^j\binom{m}{j-i}\binom{n}{i}(-1)^i0^{n-i}\tag{5}\\[9pt]
&=0\tag{6}
\end{align}
$$
Explanation:
$(1)$: Vandermonde Identity
$(2)$: rearrange terms
$(3)$: $\binom{n\vphantom{k}}{k}\!\!\binom{k}{i}=\binom{n\vphantom{k}}{i}\!\!\binom{n-i}{k-i}$ by expansion into factorials
$(4)$: rearrange terms
$(5)$: binomial expansion of $(1-1)^{n-i}$
$(6)$: all terms in the sum are zero
Any polynomial in $m$ of degree less than $n$ can be written as a linear combination of $\binom{m}{j}$ for $j\lt n$. Therefore, if $P(x)$ is a polynomial of degree less than $n$, then $(6)$ says that
$$
\sum_{k=0}^n(-1)^k\binom{n}{k}P(m+k)=0\tag{7}
$$
Since $\color{#C00000}{k^n-n!\binom{k}{n}}$ is a polynomial in $k$ of degree less than $n$,
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n
&=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\color{#C00000}{k^n}\\
&=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\color{#C00000}{n!\binom{k}{n}}\\
&=n!\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\binom{k}{n}\\[9pt]
&=n!\tag{8}
\end{align}
$$
since the only term in the last sum that is not zero is when $k=n$.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{n! = \sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\pars{n - k}^{n}}$

\begin{align}
\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\pars{n - k}^{n}
&=n!\bracks{\overbrace{\color{#c00000}{%
{1 \over n!}\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\pars{n - k}^{n}}}
^{\ds{{\rm S}\pars{n,n} = 1}}}
\end{align}
  The $\color{#c00000}{\ds{\mbox{above red expression}}}$ is the
  Stirling Number of the Second Kind
  $\ds{{\rm S}\pars{n,n}}$. See formula $\pars{10}$ in the just cited link.

$\ds{{\rm S}\pars{n,n}}$ is the number of ways of partitioning a set of $\ds{n}$ elements into $\ds{n}$ nonempty sets which is obviously $\large\tt\mbox{equal to one}$.

$$\color{#66f}{\large%
n! = \sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\pars{n - k}^{n}}
$$

A: If $(Δp)(x)=p(x)−p(x-1)$ denotes the step-one difference operator, one gets by repeated application
$$
(Δ^np)(x)=\sum_{k=0}^n\binom{n}{k} (-1)^{k} p(x-k)
$$
and your expression is equal to
$$
n!=(Δ^np)(n)
$$
where $p(x)=x^n$. 
Now in general, the degree decreases by one in each application of the difference operator, $\deg(Δ^np)=\deg(p)-n$, so that by starting with $\deg p=n$ the result is a constant which only depends on the leading coefficient of $p$. Since 
$$
Δ(x^m)=x^m-(x-1)^m=mx^{m-1}-\binom{m}2 x^{m-1}+...
$$
the leading coefficient of $Δ^kp$ is $p_n\cdot n(n-1)...(n-k+1)$ and thus the leading coefficient of $Δ^np$ is $n!\cdot p_n$.
A: First, ask yourself the following question: $\quad\displaystyle\sum_{k=0}^na^k{n\choose k}x^{n-k}~=~?\quad$ Hint: See binomial theorem.
Then apply the following two steps repeatedly: $(1).$ Differentiate both sides with respect to x, and 
$(2).$ Multiply both sides with x. Notice how, after each two steps, $\bigg(x\dfrac d{dx}\bigg)^k\circ(x-1)^n$ can be 
rewritten as $(x-1)^{n-k}\cdot P_k(x)$, where $P_k(x)$ is a polynomial of degree k in x and n. Notice also 
that $P_k(1)=n(n-1)\cdots(n-k+1)$. Try to prove these two observations by induction. Then it 
is quite evident that $P_n(1)$ will be $n!$
