Estimate $\displaystyle\int_0^\infty\frac{t^n}{n!}e^{-e^t}dt$ accurately. How can I obtain good asymptotics for $$\gamma_n=\displaystyle\int_0^\infty\frac{t^n}{n!}e^{-e^t}dt\text{ ? }$$
[This has been already done] In particular, I would like to obtain asymptotics that show $$\sum_{n\geqslant 0}\gamma_nz^n$$
converges for every $z\in\Bbb C$. 
N.B.: The above are the coefficients when expanding $$\Gamma \left( z \right) = \sum\limits_{n \geqslant 0} {\frac{{{{\left( { - 1} \right)}^n}}}{{n!}}\frac{1}{{n + z}}}  + \sum\limits_{n \geqslant 0} {{\gamma _n}{z^n}} $$
ADD Write $${c_n} = \int\limits_0^\infty  {{t^n}{e^{ - {e^t}}}dt}  = \int\limits_0^\infty  {{e^{n\log t - {e^t}}}dt} $$
We can use something similar to Laplace's method with the expansion $${p_n}\left( x \right) = g\left( {{\rm W}\left( n \right)} \right) + g''\left( {{\rm W}\left( n \right)} \right)\frac{{{{\left( {x - {\rm W}\left( n \right)} \right)}^2}}}{2}$$
where $g(t)=n\log t-e^t$. That is, let $$\begin{cases} w_n={\rm W}(n)\\
   {\alpha _n} = n\log {w_n} - {e^{{w_n}}}  \\
   {\beta _n} = \frac{n}{{w_n^2}} + {e^{{w_n}}} \end{cases} $$
Then we're looking at something asymptotically equal to $${C_n} = \exp {\alpha _n}\int\limits_0^\infty  {\exp \left( { - {\beta _n}\frac{{{{\left( {t - {w_n}} \right)}^2}}}{2}} \right)dt} $$

$$
\begin{align}
\sum_{n=1}^\infty\gamma_nz^n
&=\sum_{n=1}^\infty\int_0^\infty\frac{t^nz^n}{n!}e^{-e^t}\,\mathrm{d}t\\
&=\int_0^\infty e^{tz}e^{-e^t}\,\mathrm{d}t\\
&=\int_0^\infty e^{t(z-1)}e^{-e^t}\,\mathrm{d}e^t\\
&=\int_{1}^\infty u^{z-1}e^{-u}\,\mathrm{d}u\\
&=\Gamma(z,1)
\end{align}
$$
The Upper Incomplete Gamma Function is an entire function. According to this answer, the power series for an entire function has an infinite radius of convergence.

$\color{#C0C0C0}{\text{idea mentioned in chat}}$
$$
\begin{align}
\int_0^\infty\frac{x^{n-1}}{(n-1)!}e^{-e^x}\,\mathrm{d}x
&=\frac1{(n-1)!}\int_1^\infty\log(t)^{n-1}e^{-t}\frac{\mathrm{d}t}{t}\\
&=\frac1{n!}\int_1^\infty\log(t)^ne^{-t}\,\mathrm{d}t\\
&=\frac1{n!}\int_1^\infty e^{-t+n\log(\log(t))}\,\mathrm{d}t\\
\end{align}
$$
Looking at the function $\phi(t)=-t+n\log(\log(t))$, we see that it reaches its maximum when $t\log(t)=n$; i.e. $t_0=e^{\mathrm{W}(n)}=\frac{n}{\mathrm{W}(n)}$.
Using the estimate
$$
\mathrm{W}(n)\approx\log(n)-\frac{\log(n)\log(\log(n))}{\log(n)+1}
$$
from this answer, at $t_0$,
$$
\begin{align}
\phi(t_0)
&=-n\left(\mathrm{W}(n)+\frac1{\mathrm{W}(n)}-\log(n)\right)\\
&\approx n\log(\log(n))
\end{align}
$$
and
$$
\begin{align}
\phi''(t_0)
&=-\frac{\mathrm{W}(n)+1}{n}\\
&\approx-\frac{\log(n)}{n}
\end{align}
$$
According to the Laplace Method, the integral would be asymptotic to
$$
\begin{align}
\frac1{n!}\sqrt{\frac{-2\pi}{\phi''(t_0)}}e^{\phi(t_0)}
&\approx\frac1{n!}\sqrt{\frac{2\pi n}{\log(n)}}\log(n)^n\\
&\approx\frac1{\sqrt{2\pi n}}\frac{e^n}{n^n}\sqrt{\frac{2\pi n}{\log(n)}}\log(n)^n\\
&=\frac1{\sqrt{\log(n)}}\left(\frac{e\log(n)}{n}\right)^n
\end{align}
$$
which dies away faster than $r^{-n}$ for any $r$.

Analysis of the Approximation to Lambert W
For $x\ge e$, the approximation
$$
\mathrm{W}(x)\approx\log(x)\left(1-\frac{\log(\log(x))}{\log(x)+1}\right)
$$
attains a maximum error of about $0.0353865$ at $x$ around $67.9411$.
At least that same precision is maintained for $x\ge\frac53$.
For all $x\gt1$, this approximation is an underestimate.
