If we set $f(x):=x^x$, then the iterations of $f(x)$ follow a steadily increasing in the complexity of the function as shown by https://www.desmos.com/calculator/wpgpilljhg.
The iterations also show a very nice order to them, with their steadily increasing.
This begs me to ask the question, what is the general function $g(n,x)=f^n(x)$? Not sure what tags to use, if you know, add them and delete this part please.
I believe that your reference is wrong from $f_3(x)$ on. The correct expression is $$ f_3(x) = \left(x^{x^{1+x}}\right)^{x^{x^{1+x}}}=x^{x^{1+x}\cdot x^{x^{1+x}}}=x^{x^{1+x+x^{1+x}}} $$ I think that the best you can have is a recursive formula: the general term is $$ f_n=x^{x^{p_n}} $$ with $p_1=1$; then $$ p_{n+1}=p_n+x^{p_n} $$ so that the first terms are \begin{align} p_1=&\ 1\\ p_2=&\ 1+x\\ p_3=&\ 1+x+x^{1+x}\\ p_4=&\ 1+x+x^{1+x}+x^{1+x+x^{1+x}}\\ \vdots&\\ \end{align}