Iterated Gamma Function special value If there is a known closed form, or series representation for The Value where y=Gamma(x) intersects with y=x
$\Gamma(x) = x$
it's a value close to 3.5623822853(9)...
I stumbled across is when I iterated the Gamma function repeatedly, noticing there was a converging basin from x={.25419706972(696)...3.56238228539} where it takes on the value of y=1
$$\Gamma \Gamma \Gamma \Gamma \Gamma \Gamma ...\Gamma {x}$$
   that lower boundary is the x value where Gamma Function intersects with y=3.56238228539
Here's a screenshot from GrafEQ i used to tinker around
https://photos.google.com/album/AF1QipMW3XAKUJdRiz33pAupFhm4oxdAbkV4n-KIbXkS/photo/AF1QipP6Ql2cbBaWjTZSaUoIod2glGaP8jg132QnAvo5
I've tried looking at alternative definitions of the Gamma function, but kept ending up with Infinite Products.
   Tables of special values didn't list this value either.
I do not expect this to be very easy, since if i apply the LambertW function, i'd end up with the Omega constant, which in itself is notoriously difficult to pin down in any tractable form.
 A: The equation $\quad \Gamma(x)=x\quad$ has an infinity of roots. An obvious one is $\quad x=1.$
From WolframAlpha :

Approximates of the roots (accurate for large negative roots) can be derived from :
$$x= -n+\epsilon =\Gamma(-n+\epsilon)=\frac{\Gamma(\epsilon) }{ (-n+\epsilon)(1-n+\epsilon)(2-n+\epsilon)...(-2+\epsilon)(-1+\epsilon) }$$
$$\Gamma(\epsilon)=\frac{1}{\epsilon}-\gamma+O(\epsilon)$$
where $\gamma$ is the Euler-Masheroni constant.
$$(-n+\epsilon)^2(1-n+\epsilon)(2-n+\epsilon)...(-2+\epsilon)(-1+\epsilon)=\frac{1}{\epsilon}-\gamma+O(\epsilon)$$
In limiting to the first approximate :
$$(-n+\epsilon)^2(1-n+\epsilon)(2-n+\epsilon)...(-2+\epsilon)(-1+\epsilon)=(-1)^{n+1}n(n!)+O(\epsilon)$$
$$(-1)^{n+1}n(n!)=\frac{1}{\epsilon}-\gamma+O(\epsilon)$$
$$\epsilon\simeq \frac{1}{(-1)^{n+1}n(n!)+\gamma}$$
$$x\simeq -n+\frac{1}{(-1)^{n+1}n(n!)+\gamma} \quad\text{with condition}\quad n>>1$$

Obviously the accuracy is good only for large $n$.
For small $n$, limiting the series expansion to the first order isn't sufficient. We would have to consider the second order and even more. But the calculus becomes more arduous.
Positive roots :
$x=1$ is the first positive root.
The second positive root has no standard closed form, as far as I know.
An analytic approximate can be derived to a first order series expansion :

This must be compared to the numerical calculus of WolfralAlpha :
$$x\simeq 3.562382...$$
In order to make the above analytic approximate more accurate, we should consider the second order of series expansion and even more if necessary. But the calculus becomes more arduous.
Definitively, the numerical method of solving (Newton-Raphson or other) is much simpler than the analytic method involving limited series expansion.
