Find a Stirling-formula like for $(x!)!$ Hi working again on the gamma function I find that :
let $0<x\leq 1$ and $1\leq k\leq \infty$ then define :
$$f(x)=(x!)!,g(x)=\left(\left(x!\right)!\right)^{\frac{1}{e^{x^{k}}}}$$
Then a conjecture :
It seems $\exists k\in(1,\infty)$ such that $g(x)$ admits an asymptote as $x\to \infty$.
Then see here Trying to prove the Stirling approximation using concavity for $x\geq 1$ we can squeeze the function $f(x)$ and remains to determine some constant .
If my conjecture is true how to achieve this ?
 A: Let $a$ to be the solution of $\psi^{(0)} (x+1)=0$ (I think that you know it).
Expand $f(x)=\log[(x!)!]$ around $x=a$ and write
$$f(x)=f(a)+\Gamma (a+1)\sum_{n=2}^\infty \frac {c_n} {n!}  \, (x-a)^n$$ The first terms are
$$\color{red}{c_2}=\psi ^{(1)}(a+1) \psi ^{(0)}(\Gamma (a+1)+1)$$
$$\color{red}{c_3}=\psi ^{(2)}(a+1) \psi ^{(0)}(\Gamma (a+1)+1)$$
$$\color{red}{c_4}=3 \Gamma (a+1) \psi ^{(1)}(a+1)^2 \psi ^{(1)}(\Gamma (a+1)+1)+$$ $$\left(3 \psi
   ^{(1)}(a+1)^2+\psi ^{(3)}(a+1)\right) \psi ^{(0)}(\Gamma (a+1)+1)$$
$$\color{red}{c_5}=10 \psi ^{(1)}(a+1) \psi ^{(2)}(a+1) (\psi ^{(0)}(\Gamma (a+1)+1)+$$ $$\Gamma (a+1) \psi
   ^{(1)}(\Gamma (a+1)+1))+\psi ^{(4)}(a+1) \psi ^{(0)}(\Gamma (a+1)+1)$$
and so on for as many terms as required.
Using the expansion to $O\left((x-a)^{n+1}\right)$, computing the norm
$$\Phi_n=\int_0^1 \Big[\log[(x!)!]-\text{approximation}_{(n)}\Big]^2\,dx$$
$$\left(
\begin{array}{cc}
n & \log_{10}\big[\Phi_n\big] \\
 2 & -5.0858 \\
 3 & -5.1547 \\
 4 & -6.3509 \\
 5 & -6.9713 \\
 6 & -7.9027 \\
 7 & -8.7240 \\
 8 & -9.5957 \\
 9 & -10.456
\end{array}
\right)$$
A: As I wrote in comment, the same work can be done for
$$g(x)=(x!)!$$
$$g(x)=g(a)+\Gamma (a+1) \Gamma (a!+1)\sum_{n=2}^\infty \frac {d_n} {n!}  \, (x-a)^n$$ The first coefficients being
$$\color{red}{d_2}=\psi ^{(1)}(a+1) \psi ^{(0)}(a!+1)$$
$$\color{red}{d_3}=\psi ^{(2)}(a+1) \psi ^{(0)}(a!+1)$$
$$\color{red}{d_4}=\left(3 \psi ^{(1)}(a+1)^2+\psi ^{(3)}(a+1)\right) \psi ^{(0)}(a!+1)+$$ $$3 \Gamma
   (a+1) \psi ^{(1)}(a+1)^2 \psi ^{(0)}(a!+1)^2+3 \Gamma (a+1) \psi ^{(1)}(a+1)^2
   \psi ^{(1)}(a!+1)$$
Computing again
$$\Psi_n=\int_0^1 \Big[[(x!)!]-\text{approximation}_{(n)}\Big]^2\,dx$$
$$\left(
\begin{array}{cc}
n & \log_{10}\big[\Psi_n\big] \\
 2 & -5.0613 \\
 3 & -5.0973 \\
 4 & -6.3083 \\
 5 & -6.7855 \\
 6 & -7.7200 \\
 7 & -8.3849 \\
 8 & -9.2041 \\
 9 & -9.9228 \\
10 & -10.691 \\
11 & -11.423
\end{array}
\right)$$
that is to say
$$\log_{10}\big[\Psi_n\big] \sim -\frac {3n+13}4$$
