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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.

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  • $\begingroup$ Mathematica answers GeneratingFunction::div: The generating function does not converge. through GeneratingFunction[Binomial[n, k]^2 *(k!)^2, k, x] . $\endgroup$
    – user64494
    Jan 20, 2019 at 17:29
  • $\begingroup$ @user64494 I've edited the question so that $k$ goes from $1$ to $n$. It should converge now, but I am looking for some clean final form for the expression. $\endgroup$
    – MathsPower
    Jan 21, 2019 at 2:14
  • $\begingroup$ The body and the title of the question were substantially changed (In the original question the generating function of $ \begin{Bmatrix} n \\ k \end{Bmatrix}^2\cdot k!^2$ was asked.) without any notice. This is not a good practice. $\endgroup$
    – user64494
    Jan 21, 2019 at 9:06
  • $\begingroup$ Apologies for that. I'll keep that in mind in the future. $\endgroup$
    – MathsPower
    Jan 21, 2019 at 9:18
  • $\begingroup$ for what it's worth, the series $1, 6, 147, 6940, 536405, 62352066,\ldots$ is not on oeis.org $\endgroup$
    – Carlo Beenakker
    Jan 21, 2019 at 11:23

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