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
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 $$ P_n (x) = n!\left[ {\left( {yz} \right)^n } \right]\left( {{{e^{\,x\,y} } \over {1 - y\left( {e^{\,z} - 1} \right)}}} \right) $$
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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".