Sum involving Stirling numbers of the second kind and factorial

Question

Is it possible to give a good upper bound for the sum

$\sum\limits_{k=0}^n k!\,{n+1\brace k+1}$,

where ${n+1\brace k+1}$ is a Stirling number of the second kind.

I am aware of the relation

$\sum\limits_{k=1}^{n} {n\brace k} (x)_k = x^n$,

but I don't know how to get a good bound from this ($n^n$ is a bad bound). I'm not familiar with combinatorics, so any sort of help would be great!

Thanks in advance.

$a_n = \sum_{k=0}^n k!\,{n\brace k}$, the ordered bell number might be a helpful tool.

Is the approximation of ordered bell number $a_n \approx \frac{n!}{2(\log 2)^{n+1}}$ a good approximation?

Answer 1

This is sequence $A000629$ in $OEIS$.

In the formula section, you will notice that, in year $2002$ Benoit Cloitre proposed as a simple asymptotics $$S_n=\sum\limits_{k=0}^n k!\,{n+1\brace k+1}\sim \frac{n!}{\log ^{n+1}(2) }$$ which is in a relative error of less than $0.01$% as soon as $n >3$.