<s>[INST] Prove Stirling numbers of the second kind, $S(n,n)=1$ Stirling numbers of the second kind,

$$
S(m,n)={m \brace n}=\frac{1}{n!}\sum_{k=0}^n(-1)^k.\binom{n}{k}.(n-k)^m
$$

When $n=m$, I can easily find $S(n,n)=1$ if I set values for $n$. Ex:
$$
S(2,2)=\frac{2^2-2.1^2+0}{2!}=\frac{2}{2}=1
$$
How do i prove it for a general case ?.ie,
$$
S(n,n)=\frac{1}{n!}\sum_{k=0}^n(-1)^k.\binom{n}{k}.(n-k)^n=1
$$
So we need to prove $\sum_{k=0}^n(-1)^k.\binom{n}{k}.(n-k)^n=n!$
My Attempt:
$$
\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n=\binom{n}{0}n^n-\binom{n}{1}(n-1)^n+\binom{n}{2}(n-2)^n-\binom{n}{3}(n-3)^n+ .....\\=n^n-n.(n-1)^n+\frac{n(n-1)}{2}.(n-2)^n-\frac{n(n-1)(n-2)}{6}.(n-3)^n+ ......
$$
 [/INST] Use the old coefficient trick
\begin{eqnarray*}
[y^n]: e^{yk} = \frac{k^n}{n!}.
\end{eqnarray*}
Now the sum can be written as (reverse the index)
\begin{eqnarray*}
\sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} \frac{k^n}{n!} &=& [y^n]: \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} (e^{y})^{k} \\
&=& [y^n] : (e^y -1)^n = 1.
\end{eqnarray*}
</s>