Closed form for product of Stirling numbers of the second kind What does the following expression evaluate to:
\begin{equation}
\sum\limits_{k=1}^n \dbinom{n}{k} \cdot k! \begin{Bmatrix} n \\ k \end{Bmatrix} \cdot k! \begin{Bmatrix} n \\ k \end{Bmatrix} 
\end{equation}
We know that $k! \begin{Bmatrix} n \\ k \end{Bmatrix} = n![x^n]:(e^x-1)^k$,  where $[x^k]:f(x)$ represents the coefficient of $x^k$ in the power series for $f(x)$. I was wondering if squaring $\left(\text{i.e., } k! \begin{Bmatrix} n \\ k \end{Bmatrix} \cdot k! \begin{Bmatrix} n \\ k \end{Bmatrix}\right)$ takes us to a different power series or just to a different coefficient in the same power series? I am looking for some clean closed form. A related expression:
\begin{equation}
\sum\limits_{k=1}^n \dbinom{n}{k} \cdot k! \begin{Bmatrix} n \\ k \end{Bmatrix} 
\end{equation}
is proven to be equal to $n^n$ in this answer https://math.stackexchange.com/q/3076350.
Note: The series $1,6,147,6940,536405,62352066, \dots$ is not on oeis.org
 A: Through the Eulerian Number of 1st kind $ \left\langle \matrix{n \cr  m\cr}  \right\rangle$ we get the following identities
$$
m!\left\{ \matrix{  n \cr   m \cr}  \right\}
 = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
  \left\langle \matrix{n \cr   k \cr}  \right\rangle \left( \matrix{  k \cr   n - m \cr}  \right)}
 \quad  \Leftrightarrow \quad \left( {n - m} \right)!\left\{ \matrix{  n \cr   n - m \cr}  \right\}
 = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
  \left\langle \matrix{n \cr   k \cr}  \right\rangle \left( \matrix{  k \cr   m \cr}  \right)} 
$$
therefrom we can write our sum as
$$
\eqalign{
  & S(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
\left( \matrix{  n \cr   k \cr}  \right)k!\left\{ \matrix{  n \cr   k \cr}  \right\}k!\left\{ \matrix{  n \cr   k \cr}  \right\}}  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
 \left( \matrix{ n \cr   k \cr}  \right)k!\left\{ \matrix{  n \cr   k \cr}  \right\}
\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)}
 {\left\langle \matrix{  n \cr   j \cr}  \right\rangle \left( \matrix{  j \cr   n - k \cr}  \right)} }  =   \cr 
  &  = n!\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
 \left\{ \matrix{ n \cr   k \cr}  \right\}\left( {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)}
 {\left\langle \matrix{  n \cr   j \cr}  \right\rangle \left( \matrix{  j \cr   n - k \cr}  \right)} } \right){1 \over {\left( {n - k} \right)!}}}  \cr} 
$$
The e.g.f. for $S(n)$ then is
$$
\sum\limits_{0\, \le \,n} {S(n){{x^{\,n} } \over {n!}}}
    = \sum\limits_{0\, \le \,n} {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
 \left\{ \matrix{n \cr  k \cr}  \right\}x^{\,k} \left( {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {
 \left\langle \matrix{ n \cr   j \cr}  \right\rangle \binom{j}{n-k}
} } \right){{x^{\,n - k} } \over {\left( {n - k} \right)!}}} } 
$$
Indicating the Touchard polynomials as
$$
T_{\,n} (x) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
  \left\{ \matrix{n \cr   k \cr}  \right\}x^{\,k} } 
 = e^{\, - \,x} \sum\limits_{0\, \le \,k} {{{k^{\,n} } \over {k!}}x^{\,k} } 
$$
and the second polynomial as
$$
P_n (x) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {
 \left\langle \matrix{ n \cr   j \cr}  \right\rangle \left( \matrix{  j \cr   k \cr}  \right)} } \right){{x^{\,k} } \over {k!}}}
  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {{{\left( {n - k} \right)!} \over {k!}}\left\{ \matrix{  n \cr  n - k \cr}  \right\}x^{\,k} }  
$$
we get
$$
S(n) = n!\left[ {x^{\,n} } \right]\left( {T_{\,n} (x)P_{\,n} (x)} \right)
$$
Concerning the polynomial $P_n(x)$ let's evidence that, since
$$
{1 \over {1 - y\left( {e^{\,x}  - 1} \right)}} = \sum\limits_{0\, \le \,j} {\left( {e^{\,x}  - 1} \right)^{\,j} y^{\,j} } \quad
   = \sum\limits_{0\, \le \,k} {{{e^{\,x\,k} y^{\,k} } \over {\left( {1 + y} \right)^{\,k + 1} }}\;}
   = \sum\limits_{0\, \le \,k} {\sum\limits_{0\, \le \,j} {{{j!} \over {k!}}\left\{ \matrix{ k \cr  j \cr}  \right\}x^{\,k} y^{\,j} } } 
$$
then denoting  $P_{\, n, \, m}(x)$ as
$$
P_{n,\,m} (x) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
  \left( {n - k} \right)!\left\{ \matrix{m \cr   n - k \cr}  \right\}{{x^{\,k} } \over {k!}}} 
$$
we easily reach to
$$
\eqalign{
  & \sum\limits_{0\, \le \,m} {\sum\limits_{0\, \le \,n} {P_{n,\,m} (x)y^{\,n} {{z^{\,m} } \over {m!}}} }
    = e^{\,x\,y} \sum\limits_{0\, \le \,m} {\sum\limits_{0\, \le \,n} {{{n!} \over {m!}}\left\{ \matrix{m \cr   n \cr}  \right\}y^{\,n} z^{\,m} } }  =   \cr 
  &  = {{e^{\,x\,y} } \over {1 - y\left( {e^{\,z}  - 1} \right)}} \cr} 
$$
so that we can define $P_n(x)$ in another way as
$$
P_n (x) = n!\left[ {\left( {yz} \right)^n } \right]\left( {{{e^{\,x\,y} } \over {1 - y\left( {e^{\,z}  - 1} \right)}}} \right)
$$
---  reply to your comment   ----
Concerning the formulas used for developing $P_n(x)$, the starting point is  the basic identity
of Stirling N. 2nd kind
$$
\left\{ \matrix{  n \cr   m \cr}  \right\}\quad
  = {1 \over {m!}}\sum\limits_j {\left( \matrix{  m \cr  j \cr}  \right)j^{\,n} \left( { - 1} \right)^{\,m - j} }
  = {1 \over {m!}}\sum\limits_j {\left( \matrix{  m \cr   j \cr}  \right)\left( {m - j} \right)^{\,n} \left( { - 1} \right)^{\,j} } 
$$
then in the language of formal series
$$
\eqalign{
  & {1 \over {1 - y\left( {e^{\,x}  - 1} \right)}} = \sum\limits_{0\, \le \,j} {\left( {e^{\,x}  - 1} \right)^{\,j} y^{\,j} }  =   \cr 
  &  = \sum\limits_{0\, \le \,j} {\sum\limits_{0\, \le \,\,k\left( { \le \,j\,} \right)\,}
 {\left( { - 1} \right)^{\,j - k} \left( \matrix{  j \cr   k \cr}  \right)e^{\,xk} y^{\,j} } }  =   \cr 
  &  = \sum\limits_{0\, \le \,j} {\sum\limits_{0\, \le \,\,k\left( { \le \,j\,} \right)\,}
 {\sum\limits_{0\, \le \,l} {\left( { - 1} \right)^{\,j - k} \left( \matrix{  j \cr   k \cr}  \right){{x^{\,l} k^{\,l} } \over {l!}}y^{\,j} } } }  =   \cr 
  &  = \sum\limits_{0\, \le \,j} {y^{\,j} \sum\limits_{0\, \le \,l} {{{x^{\,l} } \over {l!}}\sum\limits_{0\, \le \,\,k\left( { \le \,j\,} \right)\,}
 {\left( { - 1} \right)^{\,j - k} \left( \matrix{  j \cr   k \cr}  \right)k^{\,l} } } }  =   \cr 
  &  = \sum\limits_{0\, \le \,j} {y^{\,j} \sum\limits_{0\, \le \,l} {{{x^{\,l} } \over {l!}}j!\left\{ \matrix{  l \cr   j \cr}  \right\}} }  \cr} 
$$
Can you follow from here ?
Concerning possible references, there are plenty concerning the properties of Stirling numbers, but each concerning some specific aspects.
A good starting point for these, and much other topics, can be the renowned "Concrete Mathematics".
