A slight variation of the famous limit definition of $e$ yields $$ \lim_{n\to\infty} \left(1+\frac 1 {n!}\right)^n =1.$$ By the comparison test, you can show that the series $$ S= \sum_{n=0}^\infty \left[ \left(1+\frac 1 {n!}\right)^n -1\right]$$ converges, and by the integral test, the corresponding integral $$ I= \int_{0}^\infty \left[ \left(1+\frac 1 {\Gamma(x+1)}\right)^x-1\right] dx $$ converges as well.
Are there closed-form expressions for the values of $S$ and $I$?
Approximate values are: $$ \begin{split} S &\approx 3.06768391298,\\ I &\approx 3.12609693980. \end{split} $$ This reminds me of the Fransén-Robinson constant, the integral of $1/\Gamma$ over $[0,\infty)$, being "almost" $e$, in the following way: $$ F=\int_0^\infty \frac{1}{\Gamma(x)}dx = \sum_{n=0}^\infty \frac{1}{n!} + \int_0^\infty \frac{e^{-x}}{\pi^2+\ln^2(x)}dx.$$ I wonder whether $I-S \approx 0.05841302682$ can also be quantified as an improper integral of some sort.