I am looking for inverse functions for the following family of functions:


$
\begin{aligned}
 f\_0(z) &= z+e^z \\
 f\_1(z) &= ze^z \\
 f\_2(z) &= z^z \\
 &\cdots \\
 f\_{n+1}(z) &= e^{\,f\_n(\log(z))} \\
\end{aligned}
$


Of course, we have $f\_1^{-1}(z) = W(z)$ with $W$ being the [Lambert W function](https://en.m.wikipedia.org/wiki/Lambert_W_function).


We also know that:


$
\begin{aligned}
f\_0^{-1}(z) &= z - W(e^z) \\
f\_2^{-1}(z) &= e^{\displaystyle W\big(\log(z)\big)} \\
\end{aligned}
$


We can get a better sense for what is going on by defining a predecessor to $W$:


$V(z) = z - W(e^z)$


This gives us:


$
\begin{aligned}
 f\_0^{-1}(z) &= V(z) \\
 f\_1^{-1}(z) &= \log(z) - V\big(\log(z)\big) \\
 f\_2^{-1}(z) &= e^{\displaystyle \log\big(\log(z)\big) - V\Big(\log\big(\log(z)\big)\Big)} \\
\end{aligned}
$


If we write $E(z) = e^z$ and $L(z) = \log(z)$, I suspect that we have:


$\,f\_{n+1}^{-1} = E^{n}\Big(L^{n+1}(z) - V\big(L^{n+1}(z)\big)\Big)$


But I have a hard time establishing it. Or my intuition is wrong and the generic inverse is something else altogether. I would be perfectly happy with a direct expression or a recurrence rule, but having both would be awesome. Also, for the time being, I am not interested in details about domains, codomains, and branches. These can be figured out later on.


Note: I suspect that the solution is trivial, but I keep running around in circles...



Generalized Lambert W Function
As per the comments, since $f\_{n+1}=E\circ f\_n\circ L$, the inverse must satisfy
$$f\_{n+1}^{-1}=L^{-1}\circ f\_{n}^{-1}\circ E^{-1}=E\circ f\_{n}^{-1}\circ L.$$
Induction gives $f\_{n}^{-1}=E^n\circ f\_0^{-1}\circ L^n$.


