Problem: Simplify the following equation:
\begin{equation}
\sum\limits_{k=1}^n \dbinom{n}{k} \begin{Bmatrix} n\\ k \end{Bmatrix} k!
\end{equation}
A solution: I am aware of the following relation:
\begin{equation}
\sum\limits_{t=1}^n t^n = \sum\limits_{k=1}^n \dbinom{n+1}{k+1} \begin{Bmatrix} n\\ k \end{Bmatrix} k!
\end{equation}
Now, I am struggling to get to a somewhat similar (i.e., clean) expression for the given equation.
Thanks for your help!
Best Answer
Using the $[x^k]:f(x)$ to represent the coeficient of $x^k$ in the power series for $f(x)$.
We recall the well known expression for the Stirling numbers of the second kind \begin{eqnarray*} \begin{Bmatrix} n\\ k \end{Bmatrix} k! = n! [x^n]:(e^x-1)^k. \end{eqnarray*} So your sum can be rewritten as \begin{eqnarray*} \sum_{k=1}^{n} \dbinom{n}{k} \begin{Bmatrix} n\\ k \end{Bmatrix} k! &=& n! [x^n]: \sum_{k=1}^{n} \dbinom{n}{k}(e^x-1)^k \\ &=& n! [x^n]: (1+e^x-1)^n -1 \\ &=& n! [x^n]: e^{nx} -1 =\color{red}{n^n}. \end{eqnarray*}