Proving that $\sum_{n=1}^{\infty}S(n,n-2)x^n = \frac{x^3(1+2x)}{(1-x)^5}$ I was wondering if anyone could give a hint on how to prove this expression, I have been stuck on it for hours. Thanks in advance!

Proving that $$\sum_{n=1}^{\infty}S(n,n-2)x^n = \frac{x^3(1+2x)}{(1-x)^5}$$ where $S(n,n-2)$ are the Stirling numbers of the second kind. 


Given the known recurrence for Stirling No. of the $2$nd kind
$$
\left\{ \matrix{
  n \cr 
  m \cr}  \right\} = m\left\{ \matrix{
  n - 1 \cr 
  m \cr}  \right\} + \left\{ \matrix{
  n - 1 \cr 
  m - 1 \cr}  \right\}
$$
we obtain
$$
\eqalign{
  & \left\{ \matrix{
  n \cr 
  n - 1 \cr}  \right\} = \left( {n - 1} \right)\left\{ \matrix{
  n - 1 \cr 
  n - 1 \cr}  \right\} + \left\{ \matrix{
  n - 1 \cr 
  n - 2 \cr}  \right\} = \left( {n - 1} \right) + \left\{ \matrix{
  n - 1 \cr 
  n - 2 \cr}  \right\}\quad  \Rightarrow \quad   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  n \cr 
  n - 1 \cr}  \right\} = \left( \matrix{
  n \cr 
  n - 2 \cr}  \right) = \left[ {0 \le n} \right]\left( \matrix{
  n \cr 
  2 \cr}  \right) \cr} 
$$
where $[P]$ is the Iverson bracket: $
\left\{ \matrix{
  [TRUE] = 1 \hfill \cr 
  [FALSE] = 0 \hfill \cr}  \right.
$
and then
$$
\eqalign{
  & \left\{ \matrix{
  n \cr 
  n - 2 \cr}  \right\} = \left( {n - 2} \right)\left\{ \matrix{
  n - 1 \cr 
  n - 2 \cr}  \right\} + \left\{ \matrix{
  n - 1 \cr 
  n - 3 \cr}  \right\} =   \cr 
  &  = \left( {n - 2} \right)\left( \matrix{
  n - 1 \cr 
  n - 3 \cr}  \right) + \left\{ \matrix{
  n - 1 \cr 
  n - 3 \cr}  \right\} = \left[ {1 \le n} \right]\left( {n - 2} \right)\left( \matrix{
  n - 1 \cr 
  2 \cr}  \right) + \left\{ \matrix{
  n - 1 \cr 
  n - 3 \cr}  \right\} \cr} 
$$
Therefrom we can write (we can take the summation index to start from $0$)
$$
\eqalign{
  & F(x) = \sum\limits_{0\, \le \,n} {\left\{ \matrix{
  n \cr 
  n - 2 \cr}  \right\}x^{\,n} }  =   \cr 
  &  = \sum\limits_{0\, \le \,n} {\left( {n - 2} \right)\left( \matrix{
  n - 1 \cr 
  n - 3 \cr}  \right)x^{\,n} }  + x\sum\limits_{0\, \le \,n} {\left\{ \matrix{
  n - 1 \cr 
  n - 3 \cr}  \right\}x^{\,n - 1} }  =   \cr 
  &  = \sum\limits_{0\, \le \,n} {\left( {n - 2} \right)\left( \matrix{
  n - 1 \cr 
  n - 3 \cr}  \right)x^{\,n} }  + x\,F(x) \cr} 
$$
which finally gives
$$
\eqalign{
  & \left( {1 - x} \right)F(x) = \sum\limits_{0\, \le \,n} {\left( {n - 2} \right)\left( \matrix{
  n - 1 \cr 
  n - 3 \cr}  \right)x^{\,n} }  = x^{\,3} \sum\limits_{0\, \le \,n} {\left( {n - 2} \right)\left( \matrix{
  n - 1 \cr 
  n - 3 \cr}  \right)x^{\,n - 3} }  =   \cr 
  &  = x^{\,3} {d \over {d\,x}}\sum\limits_{0\, \le \,n} {\left( \matrix{
  n - 1 \cr 
  n - 3 \cr}  \right)x^{\,n - 2} }  = x^{\,3} {d \over {d\,x}}{1 \over x}\sum\limits_{1\, \le \,n} {\left( \matrix{
  n - 1 \cr 
  2 \cr}  \right)x^{\,n - 1} }  = x^{\,3} {d \over {d\,x}}{1 \over x}{{x^{\,2} } \over {\left( {1 - x} \right)^{\,3} }} =   \cr 
  &  = x^{\,3} {d \over {d\,x}}{x \over {\left( {1 - x} \right)^{\,3} }} = x^{\,3} {{1 + 2x} \over {\left( {1 - x} \right)^{\,4} }} \cr} 
$$
Q.E.D.
