Inverse function of $x\mapsto \sqrt[x]x$ on $\left[0,e^{-1}\right]$ Why is it, that the inverse of $\sqrt[x]x$ is given by the infinite power tower in $x\in[\frac1e;e]$, but not in $x\in[0;\frac1e]$?
I know that the power tower diverges on that interval, but even if you always take an odd number of terms (and therefore make the tower converge), the limit is qualitatively different from the inverse of $\sqrt[x]x$. What is the deeper reason behind this?
Also, what is the inverse of $\sqrt[x]x$ in $x\in[0;\frac1e]$?
 A: If $b=\sqrt[x]x=x^\frac{1}{x}$, than x is the fixed point of $b^z$, since $b^x=x$. The fixed point is the same as the infinite power tower $b^{b^{b...}}$ if and only if the fixed point is attracting. If $x\in[\frac1e;e]$, than the fixed point is attracting, and $b\in[\exp(-e);\exp(\frac1e)]$. For $x>e$ or for positive real bases outside of this range, the fixed point is repelling, but the fixed point is still the inverse function.  For $b>\exp(\frac1e)\approx1.4447$, the fixed point becomes complex.  In the range the op asked about, if $x\in[0;e^{-1}]$, and $b=\sqrt[x]x$, the fixed point x is repelling but $x \mapsto b^{b^x}$ has two attracting fixed points, as the Op noted in stating "if you always take an odd number of terms". $b=\exp(-e)$ is the base with the 2-periodic parabolic bifurcation point.
To get the fixed point x which solves $b^x=x$, which is also the inverse function for $b=\sqrt[x]x$, one can use the Lambert-W function.  The Lambert-W function is defined as the inverse of $W e^W$. This solution works for both attracting and repelling fixed points.  
$x=-\frac{W(\ln(b))}{\ln(b)}$, where $W$ is the Lambert-W function.
