A Specific Question
Let $$\xi(z)=\frac{1}{\Gamma(z)}\sum_{n=1}^\infty \frac{\Gamma(n+z-1)}{n^n}= \int_0^1 \frac{dx}{(1+x\ln x)^z} $$
Is $\xi(1)$ irrational (this is the title question)?
Some more follow up questions
Is $\xi(z)$ demonstrably irrational for any given $z\in \mathbb{N}$? And what if we aren't asking about any specific natural number but allow for nonconstructive proofs that demonstrate $\xi$ must be irrational on at least one natural number?
Thoughts
Obviously, they're surely all irrational (because a number plucked from the reals should be irrational) but of course we are interested in demonstrably irrational. Generally we may write $$\int_{0}^{1}\frac{\left(-x\ln x\right)^{k}}{\left(1+x\ln x\right)^{z}}dx=\int_{0}^{1}\frac{1}{\left(1+x\ln x\right)^{z}}dx-\sum_{n=1}^{k}\int_{0}^{1}\frac{\left(-x\ln x\right)^{k-n}}{\left(1+x\ln x\right)^{z-1}}dx$$ But in the case $z=1$ this is simply
$$\int_0^1 \frac{(-x \ln x)^k}{1+x\ln x}dx=\sum_{n=k+1}^\infty \frac{\Gamma(n)}{n^n}$$
And $(-x \ln x)$ is maximally $1/e$ so we have some of the conditions which resemble Beuker's proof of the irrationality of $\zeta(2)$ and $\zeta(3)$ with $\zeta(s)=\sum_{n=1}^\infty n^{-s}$, the Riemann Zeta function. That is, we have an integral that we can manipulate for bounds on the tail of our series. But this type of argument works for small values of $s$ because the growth rate of $\operatorname{lcm}(1,2,\dots n)^s$ can be overcome but of course $\operatorname{lcm}(1,2, \dots n^n)$ grows even more rapidly.
On the other hand, maybe $\xi$ is better behaved than $\zeta$ in the sense that $ \xi$ is an entire function.
$\xi(2)$ is asked about on MSE and crossposted to MO. There is a pretty continued fraction form there $$\xi(2)=\frac{1}{2+\underset{n=2}{\overset{\infty}{K} }\frac{-n^{2n-1}}{n^n+(n+1)^n}}=1+\frac{1}{2+\frac{-8}{13+\frac{-243}{91+\frac{-16384}{881+\frac{-1953125}{10901+\frac{-362797056}{164305+\frac{...}{...}}}}}}}$$ I'm hopeful that this continued fraction can be generalized to $\xi(z)$.
I admit this question is near duplicate to the links above. My only defense is that $\xi(1)$ looks like a simpler thing that $\xi(2)$ so I would expect that the puzzle is easier here.
