Inverse function of $x^{x^x}$

This is not a closed form, but it actually works. It uses a strategy of solving for a variable recursively. See this graph for proof as the recursive function converges to a vertical or horizontal line as the solution is x=recursive(x). It will be easier to calculate without splitting the bases of the logarithm:

$$\mathrm{x^{x^x}=y=y_0\implies x=A_0(y_0)=log_x\left(log_x\left(y_0\right)\right)=A_1(y_0)\implies x=\lim_{n\to\infty}A_n(y_0), A_n= log_{A_n(y_0)}\left(log_{A_n(y_0)}\left(y_0\right)\right)\implies A_2(y_0)= log_{log_x\left(log_x\left(y_0\right)\right)}\left(log_{log_x\left(log_x\left(y_0\right)\right)}\left(y_0\right)\right), A_3(y_0)= log_{log_{log_x\left(log_x\left(y_0\right)\right)}\left(log_{log_x\left(log_x\left(y_0\right)\right)}\left(y_0\right)\right)}\left(log_{log_{log_x\left(log_x\left(y_0\right)\right)}\left(log_{log_x\left(log_x\left(y_0\right)\right)}\left(y_0\right)\right)}\left(y_0\right)\right),…\implies x=A_\infty (x_0)}$$

See graph 2 to see how to solve the the value that gets you y. The graph converges to a line as well like in paragraph 1:

$$\mathrm{x^{x^x}=y, B_0(a)=x=a\implies x=B_1(a)=\frac{ln(ln(a))-ln\left(ln\left(x\right)\right)}{ln\left(x\right)}\implies x=\lim_{n\to \infty} B_n(a), B_{n+1}(a)= \frac{ln(ln(a))-ln\left(ln\left(B_n(a)\right)\right)}{ln\left(B_n(a)\right)}\implies x=B_{\infty}(a)\implies B_2(a)= \frac{ln(ln(a))-ln\left(ln\left({\frac{ln(ln(a))-ln\left(ln\left(x\right)\right)}{ln\left(x\right)}}\right)\right)}{ln\left({\frac{ln(ln(a))-ln\left(ln\left(x\right)\right)}{ln\left(x\right)}}\right)}, B_3(a)= \frac{ln(ln(a))-ln\left(ln\left({ \frac{ln(ln(a))-ln\left(ln\left({\frac{ln(ln(a))-ln\left(ln\left(x\right)\right)}{ln\left(x\right)}}\right)\right)}{ln\left({\frac{ln(ln(a))-ln\left(ln\left(x\right)\right)}{ln\left(x\right)}}\right)}}\right)\right)}{ln\left({ \frac{ln(ln(a))-ln\left(ln\left({\frac{ln(ln(a))-ln\left(ln\left(x\right)\right)}{ln\left(x\right)}}\right)\right)}{ln\left({\frac{ln(ln(a))-ln\left(ln\left(x\right)\right)}{ln\left(x\right)}}\right)}}\right)},…\implies x=B_\infty(a)}$$

I will work on this a bit more, but this causes $2^n$ amounts of x’s to have to be substituted. Maybe a recursive definition using Newton’s or Euler’s rule would work. Please correct me and give me feedback!