$$\text{Question}$$
What is the general solution of the functional equation: $$ \begin{align*} f\left( x,\, y \right) &= x^{f\left( x,\, y - 1 \right)} \tag{1} \label{eq: 1}\\ \end{align*} $$ where $\left\{ x,\, y,\, f\left( x,\, y \right) \right\} \in \mathbb{C}$ and $x \ne 0$?
$$\text{Background}$$
I've recently developed an interest in functional equations of all kinds. In doing so, I fed on the solutions of partial functional equations, encountering the equation from the title. I wanted to solve $\eqref{eq: 1}$ for $f\left( x,\, y \right)$ but I only found single solutions like constants and one solution in the form of hyperoperators.
$$\text{My Thoughts}$$
$\text{Reverse Of Recursion}$
My second idea (the first idea was useless) would be to reverse the recursion by rewriting it once: $$ \begin{align*} f\left( x,\, y \right) &= x^{f\left( x,\, y - 1 \right)}\\ f\left( x,\, y \right) &= x^{x^{f\left( x,\, y - 2 \right)}}\\ f\left( x,\, y \right) &= x^{x^{x^{f\left( x,\, y - 3 \right)}}}\\ f\left( x,\, y \right) &= x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}\\ \end{align*} $$
Iff $y \to \infty$ we would get the infinit power tower, wich could be writen as: $$ \begin{align*} x &= \frac{\operatorname{W}\limits_{k}\left( \ln\left( -f\left( x,\, \infty \right) \right) \right)}{\ln\left( -f\left( x,\, \infty \right) \right)} \wedge k \in \mathbb{Z}\\ \end{align*} $$
where $\operatorname{W}$ is the Lambert W-Function.
$\text{Rewriting}$
We know:
$$ \begin{align*} f\left( x,\, y \right) &= x^{f\left( x,\, y - 1 \right)}\\ f\left( x,\, y \right) &= x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}\\ \end{align*} $$
This is a Power Tower. We can write this as $f\left( x,\, y \right) \equiv {^{y}x} = x \uparrow\uparrow y = x \uparrow^{2} y$ (see Knuth up-arrow notation) iff $y \in \mathbb{N} \cup \left\{ -1,\, 0 \right\}$. This is known as Tetration.
So we could also do the problem as a generalization of tetration for complex $y$.
$\text{Nested Function}$
Let's define ${_{n}f\left( x \right)}$ is an $n$ times iterated function $f\left( x \right)$ for $n \in \mathbb{N}$ or a superfunction for $n \in \mathbb{C}$. Now definie $e_{c}\left( x \right) = c^{x}$ is a family of functions where $c = \text{constant}$, then ${_{n}e_{c}}\left( x \right) \equiv c^{c^{\cdot^{\cdot^{\cdot^{x}}}}} \implies {_{n}e_{c}}\left( c \right) = {^{n + 1}c} = f\left( 1,\, n + 1 \right)$ or more often writen as $e_{c}^{n}\left( c \right) = f\left( c,\, n + 1 \right)$ aka $e_{c}^{n}\left( 1 \right) = f\left( c,\, n \right)$, so:
$ \begin{align*} e_{c}^{1}\left( x \right) &= c^{x}\\ e_{c}^{\frac{1}{2}}\left( e_{c}^{\frac{1}{2}}\left( x \right) \right) &= c^{x}\\\ e_{c}^{\frac{1}{3}}\left( e_{c}^{\frac{1}{3}}\left( e_{c}^{\frac{1}{3}}\left( x \right) \right) \right) &= c^{x}\\\ e_{c}^{\frac{1}{4}}\left( e_{c}^{\frac{1}{4}}\left( e_{c}^{\frac{1}{4}}\left( e_{c}^{\frac{1}{4}}\left( x \right) \right) \right) \right) &= c^{x}\\\ \end{align*} $ $\implies$ $ \begin{align*} \overbrace{e_{x}^{\frac{1}{s}}\left( \cdots \right)}^{s\, \text{times}} &= c^{x}\\\ \overbrace{e_{x}^{\frac{1}{s}}\left( \cdots \right)}^{s\, \text{times}} &= e_{x}^{1}\left( x \right)\\\ \end{align*} $
This is a clasical class of equation wich is solve'able via the Functional Square Root but there is often no closed form and this would only work for $y \in \mathbb{Q}$.
$\text{Other Stuff}$
Iff $f\left( x,\, y \right) = x \implies x = x^{x}\, \mid\, \text{not true for all}\, x$, $f\left( x,\, y \right) = x^{x} \implies x^{x} = x^{x^{x}}\, \mid\, \text{not true for all}\, x$ but there is the a pettern wich implies $f\left( x,\, y \right) = \lim\limits_{k \to \infty}\left[ {^{k}x} \right]$.