Im trying to solve for the exact solution of
$$ x^{x^x} = 17 $$
I understand that the previous solution, $x^x=17$, does not have an exact closed form solution and requires use of the Lambert W function, however i have not found an answer (from wolfram included) that seems to provide a non approximated answer even without said constraints. When approximating, however, the final answer appears to be close to $2.009$.
Does anyone know of a closed-form solution to this problem, be it even if it involves use of non-elementary functions?
If you want a definition of the exact value, consider that it is the limit, when $n\to \infty$, of the first iterate of a Newton-type method of order $n$.
Thsi will give for example, with $L=\log(2)$ $$x_{(2)}=\frac{65+128 L+128 L^2}{32+64 L+64 L^2}$$ $$x_{(3)}=\frac{274+1084 L+2136 L^2+2120 L^3+1056 L^4}{135+538 L+1064 L^2+1060 L^3+528 L^4}$$ $$x_{(4)}=\frac{ 27735+164754 L+487008 L^2+855140 L^3+951704 L^4+630240 L^5+209024 L^6 } {13665+81570 L+241908 L^2+425980 L^3+475060 L^4+315120 L^5+104512 L^6 } $$
The first decimal values $$\left( \begin{array}{cc} n & x_{(n)} \\ 2 & \color{red}{2.00}93361604829833 \\ 3 & \color{red}{2.00896}17010469369 \\ 4 & \color{red}{2.0089646}624948464 \\ 5 & \color{red}{2.0089646909}446277 \\ 6 & \color{red}{2.0089646909}923880 \\ 7 & \color{red}{2.008964690988}6512 \\ 8 & \color{red}{2.00896469098858}31 \\ \cdots & \cdots \\ \infty & \color{red}{2.0089646909885827} \end{array} \right)$$