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 $${C_n} = \exp {\alpha _n}\int\limits_0^\infty  {\exp \left( { - {\beta _n}\frac{{{{\left( {t - {w_n}} \right)}^2}}}{2}} \right)dt} $$
 A: To get an asymptotic estimate we can use a trick similar to one used to find the asymptotics of $n!$.
You wrote
$$
c_n = \int_0^\infty t^n e^{-e^t}\,dt = \int_0^\infty \exp\left(n\log t - e^t\right)\,dt = \int_0^\infty \exp f_n(t)\,dt
$$
and saw that $f_n(t)$ has a maximum at $t = W(n)$.  We thus scale the integration variable by
$$
t = W(n)(1+s),
$$
which yields
$$
\begin{align}
c_n &= W(n)^{n+1} \int_{-1}^\infty \exp\left\{n\left[\log(1+s) - W(n)^{-1} e^{W(n)s}\right]\right\}\,ds \\
&= W(n)^{n+1} \int_{-1}^\infty \exp\left\{n g_n(s)\right\}\,ds
\end{align}
$$
The largest contribution to the integral now comes from a neighborhood of $s=0$, and there we have
$$
g_n(s) = -\frac{1}{W(n)} - \frac{1+W(n)}{2}\,s^2 + \cdots.
$$
The details of the Laplace method can be worked out as usual to arrive at the conclusion that
$$
\begin{align}
c_n &\sim W(n)^{n+1} \int_{-\infty}^{\infty} \exp\left\{n\left[-\frac{1}{W(n)} - \frac{1+W(n)}{2}\,s^2\right]\right\}\,ds \\
&= W(n)^{n+1} e^{-n/W(n)} \sqrt{\frac{2\pi}{n(1+W(n))}}.
\end{align}
$$
Below is a plot of $c_n$ in blue and this asymptotic in purple for $1 < n < 10$.  Below that is a plot of $W(n)^{-n-1} e^{n/W(n)} c_n$ in blue and $\sqrt{\frac{2\pi}{n(1+W(n))}}$ in purple for $1 < n < 50$.


We can then use the estimate
$$
\log n - \log\log n < W(n) < \log n
$$
from this answer, true for $n > e$, to find that
$$
c_n < (\log n)^{n+1} e^{-n/\log n} \sqrt{\frac{2\pi}{n(1+\log n - \log\log n)}}
$$
for $n$ large enough.  Since $n! \geq (n/e)^n \sqrt{2\pi n}$ we obtain the asymptotic bound
$$
\begin{align}
\gamma_n &= \frac{c_n}{n!} \\
&< \left(\frac{e \log n}{n}\right)^{n+1} e^{-1-n/\log n} (1+\log n-\log\log n)^{-1/2}
\end{align}
$$
which clearly decreases faster than exponentially.  From this it follows that $\sum \gamma_n z^n$ converges for all $z$.
A: For $t>0$, it is true that $e^t>t^2$. Using this bound on the integral gives 
$$\int_0^\infty\frac{t^n}{n!}e^{-e^t}dt < \int_0^\infty\frac{t^n}{n!}e^{-t^2}dt.$$
The last integral is just $\frac{1}{2}$ of 
$$\frac{\Gamma\left( \frac{n+1}{2} \right)}{n!}.$$
This decays quickly enough to give convergence in the entire complex plane. 
A: $$
\begin{align}
\sum_{n=1}^\infty\gamma_nz^n
&=\sum_{n=1}^\infty\int_0^\infty\frac{t^nz^n}{n!}e^{-e^t}\,\mathrm{d}t\\
&=\int_0^\infty e^{tz}e^{-e^t}\,\mathrm{d}t\\
&=\int_0^\infty e^{t(z-1)}e^{-e^t}\,\mathrm{d}e^t\\
&=\int_{1}^\infty u^{z-1}e^{-u}\,\mathrm{d}u\\
&=\Gamma(z,1)
\end{align}
$$
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.
