I have a problem simplifying the summation here:

$$ \sum_{x=0}^{n}\sum_{y=0}^{x} {n\choose{x}} {x\choose{y}} y!(x-y)! $$

The last three terms can be simplified to x!, so the current summation becomes:

$$ \sum_{x=0}^{n}\sum_{y=0}^{x} {n\choose{x}} x! = \sum_{x=0}^{n}{n\choose{x}}(x+1)! $$

Then I get stuck here, it seems it should relate to $e$, please inspire me if any one knows the answer.

Thanks for the help.