Some month ago and after several attempt to find it I have :
Let $y\geq 1$
$$e^{y}\overset{?}{=}\lim_{x \to 0^+}\left(\left|1-\frac{y-1+x!}{1-\frac{1}{1-x!x!!x!!!x!!!!x!!!!!\cdot\cdot\cdot}}\right|\right)^{\frac{1}{x}}$$
Where we have the composition of Gamma function ($x!!=\Gamma(\Gamma(x+1)+1)$)
As Robjohn answered a previous question Conjecture: $\lim\limits_{x\to 0}(x!\,x!!\,x!!!\,x!!!!\cdots )^{-1/x}\stackrel?=e$ .I have tried the same path without any success .
If true might be this equality have some nice application .
Also I wondering myself if it could be true if $y$ is a complex number .
Some matter :
It seems there is again the binomial coefficient which appears .
Let :
$$f_n(x)=\left(\left|1-\frac{y-1+x!}{1-\frac{1}{1-x!x!!x!!!x!!!!x!!!!!...x!!!...!}}\right|\right)^{\frac{1}{x}}$$
Where we have composing $n$ times the Gamma function with itself .
Then it seems we have :
$$\lim_{x\to 0^+}f_n(x)=e^{y(1-(1-\gamma)^n)}$$
As attempt we can try to use log as the limit is also :
$$\lim_{x\to 0^+}\frac{\ln\left(1-\frac{\left(y-1+x!\right)}{1-\frac{1}{1-x!x!!x!!!\cdot\cdot\cdot}}\right)}{x}=y$$
And then use the linked question with Robjohn's answer.The fraction form can be simplified .
Edit 24/01/2023 :
There is another way if :
$$\lim_{x \to 0^+}\left(\left|1-\frac{y-1+x!}{1-\frac{1}{1-x!x!!x!!!x!!!!x!!!!!\cdot\cdot\cdot}}\right|\right)^{\frac{1}{x}}=\lim_{x \to 0^+}\left(\left|1-\frac{y-1+x!x!!x!!!x!!!!x!!!!!\cdot\cdot\cdot}{1-\frac{1}{1-x!x!!x!!!x!!!!x!!!!!\cdot\cdot\cdot}}\right|\right)^{\frac{1}{x}}$$
Then we can write :
$$\lim_{x \to 0^+}\left|g(x!x!!x!!!x!!!!x!!!!!\cdot\cdot\cdot)\right|^{\frac{1}{x}}\overset{?}{=} e^y$$
Where :
$$g(x)=1-\frac{y-1+x}{1-\frac{1}{1-x}}=\frac{y-1}{x}-x-y+3$$
How to (dis)prove it ?
