Asymptotic Behaviour of $\Gamma^{(k)}(1)$ Pretty simple question to which I haven't found much in the literature. Are there proper results for the asymptotic behaviour of the derivatives of the Gamma-function? That is for
$$\Gamma^{(k)}(1) = \int_0^\infty (\log t)^k e^{-t} \, {\rm d}t$$
as $k$ gets large?

The range of integration over the interval $(0,1)$ is responsible, for the strong factorial growth which I oversaw. Hence, I'd be somewhat more interested in the asymptotics of 
$$\int_1^\infty (\log t)^k e^{-t} \, {\rm d}t \, .$$
The integrand has a maximum at approximately $t\approx \frac{k}{\log k}$, so it will blow up somehow. Laplace's method is what comes to my mind, which gives
$$\int_1^\infty (\log t)^k e^{-t} \, {\rm d}t \sim \sqrt{\frac{2\pi k}{W(k)+1}} \, W(k)^k \, e^{-\frac{k}{W(k)}}$$
where $W(k)$ is the principal branch of LambertW.
The remainder for the asymptotic expansion of the starting expression (given by metamorphy) is then just
$$\frac{\int_1^\infty (\log t)^k e^{-t} \, {\rm d}t}{k!} \sim \frac{1}{\sqrt{W(k)+1}} \left( \frac{e \, W(k)}{k \, e^{\frac{1}{W(k)}}} \right)^k \, ,$$
vanishing faster than any power.
Is it possible to obtain an asymptotic expansion, instead of just the asymptotics?
 A: $\Gamma(1+z)$ has simple poles at $z=-n$ with the residue $(-1)^{n-1}/(n-1)!$ where $n$ takes positive integer values. Thus, for any positive integer $m$, the function $$\Gamma(1+z)+\sum_{n=1}^{m}\frac{(-1)^n}{(n-1)!}\frac{1}{n+z}=\sum_{k=0}^{\infty}\left(\frac{\Gamma^{(k)}(1)}{k!}+(-1)^k\sum_{n=1}^{m}\frac{(-1)^n}{n^k\cdot n!}\right)z^k$$ is regular in $|z|<m+1$, and the last series converges at $z=m$ (at least). This gives the asymptotics $$\Gamma^{(k)}(1)\asymp(-1)^k k!\sum_{n=1}^{(\infty)}\frac{(-1)^{n-1}}{n^k\cdot n!},\qquad k\to\infty.$$ Despite the convergence, this is only an asymptotic equality; there is a remainder that is not captured. (It is coming from $\int_1^\infty t^z e^{-t}\,dt$ which is an entire function of $z$.)
A: Note that $$\sum_{k=0}^\infty \Gamma^{(k)}(1) \frac{z^k}{k!} = \Gamma(1+z)$$
The closest singularity of $\Gamma(\zeta)$ to $\zeta=1$ is at $\zeta=0$, where $\Gamma$ has a simple pole with residue $1$. The next closest singularity is at $\zeta=-1$. Thus 
$$\Gamma(1+z) - \frac{1}{1+z} = \sum_{k=0}^\infty \left(\frac{\Gamma^{(k)}(1)}{k!} - (-1)^{k}\right) z^k $$
has radius of convergence $2$.  Thus $\Gamma^{(k)}(1) \sim (-1)^k k!$,
with
$$ \left| \Gamma^{k}(1) - (-1)^k k! \right| = O\left(r^k k!\right)\ \text{for all } r \in (0,1/2)$$
