prove $\sum_{r=1}^{n-1}(n-r)^2\binom{n-1}{n-r}=n(n-1)2^{n-3}$ This is problem 2.38 from the book "Principles and Techniques in Combinatorics". 
Prove $$\sum_{r=1}^{n-1}(n-r)^2\binom{n-1}{n-r}=n(n-1)2^{n-3}$$
I have have been trying different things on and off for two days and still can't derive RHS from LHS.
what I have got so far:
$$\sum_{r=1}^{n-1}(n-r)^2\cdot\frac{n-1}{n-r}\binom{n-2}{n-r-1}
=(n-1)\sum_{r=1}^{n-1}(n-r)\cdot\binom{n-2}{n-r-1}
=\\(n-1)\sum_{r=2}^{n}(n-r)\cdot\binom{n-2}{n-r}
=(n-1)\sum_{r=2}^{n}(n-r)\cdot\binom{n-2}{n-r}
=\\(n-1)\sum_{r=2}^{n}(n-r)\cdot\frac{n-2}{n-r}\binom{n-3}{n-r-1}
=\\(n-1)(n-2)\sum_{r=2}^{n}\cdot\binom{n-3}{n-r-1}
$$
any hints or solutions would be greatly appreciated. 

Another way:
$$
\eqalign{
  & \sum\limits_{1\, \le \,r\, \le \,n - 1} {\left( {n - r} \right)^{\,2} \left( \matrix{
  n - 1 \cr 
  n - r \cr}  \right)}  = \sum\limits_{1\, \le \,k\, \le \,n - 1} {k^{\,2} \left( \matrix{
  n - 1 \cr 
  k \cr}  \right)}  =   \cr 
  &  = \sum\limits_{0\, \le \,k\, \le \,n - 1} {k^{\,2} \left( \matrix{
  n - 1 \cr 
  k \cr}  \right)}  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n - 1} \right)} {\left( {k\left( {k - 1} \right) + k} \right)\left( \matrix{
  n - 1 \cr 
  k \cr}  \right)}  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n - 1} \right)} {k\left( {k - 1} \right)\left( \matrix{
  n - 1 \cr 
  k \cr}  \right)}  + \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n - 1} \right)} {k\left( \matrix{
  n - 1 \cr 
  k \cr}  \right)}  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n - 1} \right)} {k\left( {k - 1} \right){{\left( {n - 1} \right)^{\,\underline {\,k\,} } } \over {k!}}}
  + \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n - 1} \right)} {k{{\left( {n - 1} \right)^{\,\underline {\,k\,} } } \over {k!}}}  =   \cr 
  &  = \left( {n - 1} \right)\left( {n - 2} \right)\sum\limits_{2\, \le \,k\, \le \,n - 1} {{{\left( {n - 3} \right)^{\,\underline {\,k - 2\,} } } \over {\left( {k - 2} \right)!}}}
  + \left( {n - 1} \right)\sum\limits_{1\, \le \,k\, \le \,n - 1} {{{\left( {n - 2} \right)^{\,\underline {\,k - 1\,} } } \over {\left( {k - 1} \right)!}}}  =   \cr 
  &  = \left( {n - 1} \right)\left( {n - 2} \right)\sum\limits_{2\, \le \,k\, \le \,n - 1} {\left( \matrix{
  n - 3 \cr 
  k - 2 \cr}  \right)}  + \left( {n - 1} \right)\sum\limits_{1\, \le \,k\, \le \,n - 1} {\left( \matrix{
  n - 2 \cr 
  k - 1 \cr}  \right)}  =   \cr 
  &  = \left( {n - 1} \right)\left( {n - 2} \right)2^{n - 3}  + \left( {n - 1} \right)2^{n - 2}  =   \cr 
  &  = n\left( {n - 1} \right)2^{n - 3}  \cr} 
$$
