# observation from a question on m.se

Set $g(z)=\sum_{n=1}^{\infty} \left(\frac{z}{n}\right)^{n}$, then the sum:

$\sum_{m=1}^{\infty} \frac{m!}{m^{m}}$ can be interpreted as the value of the Laplace

Transform of $g(z)$ at $s=1$:

$\sum_{m=1}^{\infty} \frac{m!}{m^{m}} = \sum_{m=1}^{\infty} \frac{1}{m^{m}} \int_{0}^{\infty} t^{m}e^{-t} dt = \int_{0}^{\infty} e^{-t} \sum_{m=1}^{\infty} \frac{t^{m}}{m^{m}} dt$