If $u(x)=x^x$ then we can form $$ u^2(x) = \left(x^x \right)^{x^x} = x^{x^{x+1}} $$ some simplification occurs, but the further iterates are a typographical challenge to mathjax.
Writing $E_k$ for $u^k(x)$ we have $$ E_0 =x \\ E_{n+1} = u(E_n) = E_n^{E_n}. $$ However, each term in the sequence is a power of $E_0 = x$ so we may also write $$ E_n = x^{P_n} $$ where $$\begin{align} P_0 &= 1\\ P_1 &= x \\ P_2 &= x^{x+1} \end{align} $$ the iteration gives $$ E_{n+1} = E_n^{E_n} = \left(x^{P_n}\right)^{x^{P_n}} = x^{P_nx^{P_n}} $$ so that $$ P_{n+1}=P_n x^{P_n}. $$ We note that $P_k$ is also a power of $x$, say $P_n=x^{Q_n}$ which gives $$ P_{n+1} = x^{P_n+Q_n} $$ it seems that each iteration takes us one step further up the ladder of tetration, but i do not know the notation to express this in a precise symbolic form. can anyone help?
Extending my earlier comment. First we write $f(x)=x^x$ and $f(x,h)=f(f(x,h-1))$ , $f(x,0)=x$ and $f(x,1)=f(x)$ where $h$ indicates the "iteration-height". Then we consider the function $g(x,h)=f(x+1,h)-1$ and get a nicely iterable power series for $g(x)$ . The coefficients of $g(x,h)$ depend on $h$ and are in fact polynomials in $h$ of increasing order. The first few coefficients are $$ \begin{array} {rrrlrrrr} g(x,h) &= & x \cdot & (1 &&&&) \\ & +& x^2 \cdot & (&1 \cdot h&&&) \\ & +& x^3 \cdot & (&-1/2 \cdot h & + 1 \cdot h^2&&) \\ & +& x^4 \cdot & (& 7/12 \cdot h & -5/4 \cdot h^2 & + 1 \cdot h^3&) \\ & \vdots & \vdots \end{array}$$ After that, the actual evaluation to any arbitrary iteration-height (I think: even fractional height) can be done by evaluating $g(x)$ at $x-1$ because $$f(x,h)=g(x-1,h)+1$$
However, the next step must be to determine the range of convergence by analyzing the pattern of the coefficients in the polynomial to determine the "general term" for the polynomials in $g(x,h)$.
The convergence-radius might easily be zero, especially for the fractional iterates, but if this is the case then possibly a summation-method for divergent series can still help.