Iteration of $x \to x^x$ 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?
 A: 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.
