Another strange limit which seems to converges to $\gamma$ Hi in trying to show that the Euler Mascheroni constant is irrational or not I find empiricaly :
$$\lim_{n\to\infty,x\to 0}f_n(x)=\lim_{n\to\infty,x\to 0}\frac{x!!!...!^{x!!...!^{x!...!^{...^{x!}}}}-x!^{x!!^{x!!!^{...^{x!!!...!}}}}}{x}=^?\gamma=0.5772...$$
Some explanation :
In the case $n=2$ we have :
$$f_2(x)=\frac{x!!^{x!}-x!^{x!!}}{x}$$
In the case $n=3$ we have :
$$f_3(x)=\frac{x!!!^{x!!^{x!}}-x!^{x!!^{x!!!}}}{x}$$
And so on...
On the other hand we have :
$$x!=\Gamma(x+1),x!!=\Gamma(\Gamma(x+1)+1)...$$
As clue we have :
Let :
$g(x)=\frac{d}{dx}x!$
Then :
$$-g(0)=\gamma$$
When $k=2n$ ,$n\geq 3$ an integer  it seems we have :
$$\lim_{x\to 0}f_k=(-1+(1-\gamma)^{k-1})\gamma$$
And :
$$\lim_{x\to 0}\frac{\left(x!!\right)^{ax!}-\left(x!\right)^{ax!!}}{x}=\gamma^2a$$
So there is a possible induction here .
How to (dis)prove it ?
 A: If we define
$$t_0=x \qquad \text{and} \qquad t_n=(t_{n-1})!$$ Expanding around $x=0$
$$t_n=1+\sum_{k=1}^\infty a_{n,k}\, x^k$$ So, my idea is to expand
$$(t_n)^{t_{n-1}}\qquad \text{and}\qquad (t_{n-1})^{t_n}$$ as series, going first to logarithms, to continue for any level of exponentiation and then use $u=e^{\log(u)}$.
But, we just need to consider the first term and very clear patterns appear using
$$t_n=1+a_n x+O(x^2)$$ For example
$$f_4=\frac{(1+a_4 x)^{(1+a_3 x)^{(1+a_2 x)^{(1+a_1 x)}}}-
(1+a_1   x)^{(1+a_2 x)^{(1+a_3 x)^{(1+a_4 x)}}}}x$$
$$f_4=(a_4-a_1)-a_1a_2a_3a_4\, x+O(x^2)$$ and, in a general manner
$$\color{blue}{f_n=(a_n-a_1)-x \prod_{k=1}^n a_k+O(x^2)}$$
Here appears the simplicity of the difference of the first coefficients which gives
$$\left(
\begin{array}{cc}
\color{red}{\large n} & \color{red}{\large\gamma^{-2}\,(a_n-a_1)} \\
 2 & 1 \\
 3 & 2-\gamma  \\
 4 & 3-3 \gamma +\gamma ^2 \\
 5 & 4-6 \gamma +4 \gamma ^2-\gamma ^3 \\
 6 & 5-10 \gamma +10 \gamma ^2-5 \gamma ^3+\gamma ^4 \\
 7 & 6-15 \gamma +20 \gamma ^2-15 \gamma ^3+6 \gamma ^4-\gamma ^5 \\
8 & 7-21 \gamma +35 \gamma ^2-35 \gamma ^3+21 \gamma ^4-7 \gamma ^5+\gamma^6 \\
9 &8-28 \gamma +56 \gamma ^2-70 \gamma ^3+56 \gamma ^4-28 \gamma ^5+8\gamma ^6-\gamma ^7 \\
10 & 9-36 \gamma +84 \gamma ^2-126 \gamma ^3+126 \gamma ^4-84 \gamma ^5+36 \gamma ^6-9 \gamma ^7+\gamma ^8\\
\end{array}
\right)$$ which are,  as you wrote,
$$\lim_{x\to 0}f_n=\left(1-(1-\gamma)^{n-1}\right)\,\gamma$$
$$\lim_{x\to 0}(f_n-\gamma)=-\left((1-\gamma)^{n-1}\right)\,\gamma$$ which converges quite fast since
$$\frac {f_{n+1}-\gamma}{f_{n}-\gamma}=1-\gamma$$
Notice that using instead
$$t_n=1+a_n x+b_n x^2+O(x^3)$$ we should have
$$\color{blue}{f_n=(a_n-a_1)+x \left((b_n-b_1)- \prod_{k=1}^n a_k\right)+O(x^2)}$$
