Simplify the following: 


\begin{equation}
\sum\limits\_{\ell =1}^n P(n,\ell) (e^x -1)^\ell
\end{equation}


to something like $n!n^n$. I got curious about this expression after going through this answer: math.stackexchange.com/q/3076350. 



Sum of: k permutations of n $\times e^x$
$$\sum\limits\_{l=1}^n {n! \over (n-k)!} (e^x -1)^l =$$


$$e^{\frac{1}{e^x-1}} \left(e^x-1\right)^n \Gamma \left(n+1,\frac{1}{-1+e^x}\right)-1$$


