I try to simplify this limit :
$$\lim_{x\to 0} \left(2^{1-x!}3^{1-x!!}4^{1-x!!!}5^{1-x!!!!}6^{1-x!!!!!}\cdots\right)^{\frac{1}{x}}=L$$
Where we compose the Gamma function with itself .
From the past I already know that it involves Euler-Mascheroni constant (derivative of the Gamma function near zero) and it converges .
Interestingly it involves power of prime numbers and perhaps some number theory .
For example :
$$\lim_{x\to 0}\frac{\ln\left(2^{\left(1-x!\right)}3^{\left(1-\left(x!\right)!\right)}\right)}{x}=\gamma(-\gamma\ln(3)+\ln(6))$$
How to simplify it or in other word does L admits a closed form ?
The calculations are not so complicated as I thought.
Let $$t_1=x \qquad \text{and} \qquad t_n=[t_{n-1}]!\qquad \text{and} \qquad u_n=\frac 1x \,\log \left(n^{1-t_n}\right)$$ Using series expansion around $x=0$ gives at the limit $$u_n=\gamma\,(1 -\gamma)^{n-2}\, \log(n)$$ Then, the partial sum of the logarithm of the expression is $$S_p=\gamma \sum_{n=2}^p (1 -\gamma)^{n-2}\, \log(n)$$
There is an explicit result for the partial sum $$\color{blue}{S_p=\gamma (1-\gamma )^{p-1} \Phi^{(0,1,0)}(1-\gamma ,0,p+1)-\gamma \Phi^{(0,1,0)}(1-\gamma ,0,2)}$$ where $\Phi(.)$ is the Lerch transcendent function.
$$\large \color{red}{\log(L)=\underset{p\to \infty }{\text{limit }}S_p=-\gamma \, \Phi^{(0,1,0)}(1-\gamma ,0,2)}$$
which is also $$\large \color{red}{\log(L)=-\frac{\gamma }{(1-\gamma )^2}\,\,\,\text{Li}^{(1,0)}_0(1-\gamma )}$$
Numerically, $$\log(L)=0.941986151767766605347692845869094706057514927\cdots$$
$$\text{Li}^{(1,0)}_0(1-\gamma )=-0.291705209103709537352092057571457\cdots$$
Edit
Making the problem more general $$t_1=x \qquad \text{and} \qquad t_n=[t_{n-1}]!\qquad \text{and} \qquad u_n=\frac 1x \,\log \left((n+k)^{1-t_n}\right)$$ $$\large \color{red}{\log\big[L(k)\big]=-\gamma \, \Phi^{(0,1,0)}(1-\gamma ,0,2+k)}$$ which is valid for any $k>0$.
This is a very nice function of $k$.
If $k$ is an integer $(k>0)$, the limit can also write in terms of the derivative of the polylogarithm function $$\large \color{red}{\log(L_k)=-\frac{\gamma}{ (1-\gamma )^{k+2}}\,\Bigg[\text{Li}^{(1,0)}_0(1-\gamma )+\sum_{i=2}^{k+1} (1-\gamma )^i\log(i) \Bigg]}$$
Update
As a continuous function $$L(k)=\exp\left(-\gamma \,\, \Phi^{(0,1,0)}(1-\gamma ,0,2+k) \right)$$ is extrely strange : it is perfectly continuous for $k \geq -2$ and it still exists for $k<-2$ with discontinuities at each negative integer values of $k$. Between two negative integer numbers $(-6 \leq k \leq -3)$, it goes through a maximum. For $k <-6$, the function starts to be continuous again.
What is curious is the approximation $$L(k) \sim k+10 \gamma \left(e^{\gamma }-\text{Ei}(1) \log (2)\right)$$