Nested Gamma functions and $\lim_{n\to\infty} a_n=1$

Question

Hi it's a problem that I cannot prove :

Let $1\leq a_0\leq 3$ and defines the following sequence : $$\Gamma(a_n)=a_{n+1}$$ Where $\Gamma(x)$ is the Gamma function and $n\geq 0$ a natural number.

Then : $$\lim_{n\to\infty} a_n=1$$

I think we can consider two cases when $1\leq a_0\leq 2$ and $2\leq a_0\leq 3$

I have tried to use the approximated inverse Gamma function without success . Furthermore on wiki we have (result due to Kečkić and Vasić) :

$$x^{x-1}e^{1-x}<\Gamma(x)<x^{x-\frac{1}{2}}e^{1-x}$$ Where $x>1$

So if we define for $1\leq b_0\leq 2$ the following sequence :

$$b_{n+1}=b_n^{b_n-\frac{1}{2}}e^{1-b_n}$$

We see numerically that it tends to one .

But it doesn't work so well with the LHS , the $b_n$ looks complicated and we lost the initial inequality .

I have not try an experimentation with the integral definition of the Gamma function .

I think it's a hard nut so if you have just an approach to begin.

Any helps is greatly appreciated .

Thanks a lot for your contributions