inverse function of $f(x)=x^{x^x}$ I have easily found the inverse of $f(x)=x^x$ using the following:
$$y=x^x\Rightarrow \ln(y)=x\ln(x)\Rightarrow W(\ln(y))=\ln(x)\therefore x=e^{W(\ln(y))}$$
However I am struggling to do the same for $f(x)=x^{x^x}$, this is what I have tried:
$$x=y^{y^y}$$
$$\ln(x)=y^y\ln(y)$$
and if: $u=y^y$ then:
$$\frac{ue^u}{e^{W(ln(u))}}=\ln(x)$$
now by letting $\ln(u)=ve^v$ we get:
$$e^{ve^v}v=\ln(x)\Rightarrow \ln(v)+ve^v=\ln(\ln(x))$$
However I cannot see an easy way of getting from here or even if this is the correct way to approach it, as I have tried several methods and they have all failed

The closest progress I have made is if i define a function $g(x)=\ln(x)+xe^x$ then $G(x)=g^{-1}(x)$ then I get:
$$y=\exp\left(W\left(G(\ln(\ln(x)))+\ln(G(\ln(\ln(x)))\right)\right)$$
However this is a fairly ugly expression and I still have no way of defining $G(x)$
 A: If anything, I'd say perhaps we should flip this around. The functions
$$f_1(x) := x, f_2(x) := x^x, f_3(x) := x^{x^x}, \cdots$$
form a logical progression, and so too their inverses. Hence, it "seems" like that instead of scratching our head regarding how to understand the inverse of $f_3$ as a Lambert W-function because $f_2$ can be given as such, we should perhaps be more inclined to think of the Lambert W-function as the one that needs special expressing - in terms of the inverse of $f_2$! And indeed you can do that: call the $n$th inverse term from the above a "$n$-th order tetrational root", or $\mathrm{trt}_n(x)$, so that $f_2^{-1} = \mathrm{trt}_2$. Then we have, just as
$$\mathrm{trt}_2(x) = e^{W(\ln(x))}$$
which is what gets you wondering, that we can turn the tables to get
$$W(x) = \ln(\mathrm{trt}_2(e^x))$$
and so the equation that defines $W(x)$ is just seen to be a funny case involving 2nd-order tetrational roots, so when you ask to express $\mathrm{trt}_3(x)$ in terms of $W(x)$, you are essentially asking to express $\mathrm{trt}_3(x)$ in terms of $\mathrm{trt}_2(x)$, and if you think about that, that would be kind of like asking "how can I express a cube root $\sqrt[3]{\cdots}$ using square roots (and arithmetic) $\sqrt{\cdots}$ only?" which clearly isn't something you can do - so why should that change here? Moreover, if we put things this way, you see we probably wouldn't think we needed to; it's just that the sequence of "tetrational roots" is not one that is usually put in your algebraic toolkit.
Moreover, your identity involving what you call "$G(x)$" would seem to show what would be the more "correct" relation between $\mathrm{trt}_3$ and $W$: while the equation that defines $W$, i.e.
$$xe^x = a$$
will be solved using $\mathrm{trt}_2$, the equation
$$xe^x + \ln x = a$$
is the analogous one you would solve using $\mathrm{trt}_3$. If you like, it would define a sort of "higher order" $W$-function.
