Asymptotic behaviour of $\int_0^{\infty } x^{-x} \exp (n x) dx$

Using the saddle point method http://dlmf.nist.gov/2.4.iv with $a=0$, $b=+\infty$, $t_0=e^{-1}$, $z=e^n$, $p(t)=t\log t$ and $q(t)=1$, we find $$ \int_0^{ + \infty } {x^{ - x} e^{nx} dx} = e^n \int_0^{ + \infty } {e^{ - e^n t\log t} dt} \\ \sim \exp \left( {\tfrac{{n - 1}}{2} + e^{n - 1} } \right)\sqrt {2\pi } \left( {1 + \sum\limits_{k = 1}^\infty {\sqrt {\frac{{2e}}{\pi }} \Gamma \left( {k + \tfrac{1}{2}} \right)\frac{{b_{2k} }}{{e^{nk} }}} } \right), $$ where the coefficients are given as complex residues: $$ b_{2k} = \frac{1}{2}\mathop{\operatorname{Res}}\limits_{t = e^{ - 1} }\left[ {(t\log t + e^{ - 1} )^{ - k - 1/2} } \right] = \frac{1}{2}e^{k - 1/2} \mathop{\operatorname{Res}}\limits_{s = 0} \left[ {((1 + s)\log (1 + s) - s)^{ - k - 1/2} } \right]. $$ For example, $$ b_2 = - \frac{1}{{12}}\sqrt {\frac{e}{2}} ,\quad b_4 = - \frac{{23e}}{{864}}\sqrt {\frac{e}{2}} , $$ whence $$ \int_0^{ + \infty } {x^{ - x} e^{nx} dx} \sim \exp \left( {\tfrac{{n - 1}}{2} + e^{n - 1} } \right)\sqrt {2\pi } \left( {1 - \frac{1}{{24}}\frac{1}{{e^{n - 1} }} - \frac{{23}}{{1152}}\frac{1}{{e^{2(n - 1)} }} + \cdots } \right). $$