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Pat Devlin
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Expanding on some of this...

I suppose it makes no sense unless $x$ is between 0 and 1. And in this case...

Define $y =x^{x^{x^{\cdots}}}$. Then we have $y= x^y$. For any fixed $0<x<1$, this defines a unique positive value for $y$ (intermediate value theorem will imply $y$ is between 0 and 1).

So that's one way to define it in the range we want. It seems to me that solving the above equation gives us:

$$x = e^{\log(y)/y} = y^{1/y}$$.

So your function is the inverse of this. Might be related to the Lambert W function (didn't think).

Pat Devlin
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