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: $$ \sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n=\binom{n}{0}n^n-\binom{n}{1}(n-1)^n+\binom{n}{2}(n-2)^n-\binom{n}{3}(n-3)^n+ .....\\=n^n-n.(n-1)^n+\frac{n(n-1)}{2}.(n-2)^n-\frac{n(n-1)(n-2)}{6}.(n-3)^n+ ...... $$
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*}