Parametrization of the curve $x^{x^y}=y$ I was looking at the graph of the equation $x^{x^y}=y$ (Desmos link).
This graph has two components that cross at the point $(1/e^e,1/e)=(e^{-e},e^{-1})$.
Component 1 (as I'll call it) is the component $x^y=y$ which has the simple parametrization
$$(x,y)=\left(t^{1/t},t\right),\qquad0<t<\infty.$$
Component 2 is a path between the points $(0,0)$ and $(0,1)$.

Does component 2 also admit a parameterization?

To clarify: Component 2 is a path so of course it abstractly admits a parameterization, but I'm asking if there is a parametrization that we can actually write down algebraically in terms of elementary functions.

My motivation for this question is from the limiting behavior of the sequence $0,1,x,x^x,x^{x^x},x^{x^{x^x}},\ldots$, whose behavior is closely related to the solutions to $x^{x^y}=y$. In particular, if $x$ is less than $e^{-e}$ then this sequence alternates between the upper and lower parts of component 2.
 A: Edit: In the comments, Maxim found a better way to get this solution. Here's Maxim's way in detail: Start with the equation
$$x^{(x^y)}=y.$$
Raising both sides to the $y$th power gives the equation
$$(x^y)^{(x^y)}=y^y.$$
In other words, if we set $z=x^y$ then we are looking for solutions to the equation $z^z=y^y$.
There is the trivial solution $y=z$ which corresponds to component 1.
We are interested in parametrizing the nontrivial solutions.
We can rearrange the equation $z^z=y^y$ to get
$$(1/z)^{1/y}=(1/y)^{1/z}.$$
This has the well-known parametrization $1/y=t^{1/(t-1)}$ and $1/z=t^{t/(t-1)}$ (see this answer).
Then $y=t^{1/(1-t)}$ and $z=t^{t/(1-t)}$.
Finally, solving for $x$ gives
$$x=z^{1/y}=t^{(t^{-t/(1-t)})/(1-t)}.$$
We obtain the parametrization
$$\boxed{(x,y)=(t^{(t^{-t/(1-t)})/(1-t)},t^{t/(1-t)}),\qquad0<t<\infty}.$$
Interestingly, the substitution $t\leftrightarrow\frac{1}{t}$ gives the similar parametrization
$$\boxed{(x,y)=(t^{(t^{-t/(1-t)})/(1-t)},t^{1/(1-t)}),\qquad0<t<\infty}.$$
