A friend who tutors high school math showed me this equation. I could not solve it. But by trial & error in Python I discovered $\sqrt[5]{5}$ is the answer. Further realized that for any integer $n$, $x^{x^n} = n$ is solved by $\sqrt[n]{n}$. How does one handle such equations by ordinary algebraic methods?
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$\begingroup$Suppose $x^{x^n}=n$.
$$\log\left(x^{x^n}\right)=x^n\log x=\log n$$
Take the substitution $\log x=y$ (or $x=e^y$).
$$\left(e^{y}\right)^ny=ye^{yn}=\log n$$
$$\left(e^{y}\right)^ny=yne^{yn}=n\log n$$
Let $m=e^n$, then $\log n=m$
$$yne^{yn}=me^m$$
As the function $y=xe^x$ is injective in the interval $[-1,\infty)$ (which can be seen easily on a graph), if we make the assumption $m\ge-1$ (i.e. $n\ge e^{-1}$), then we have
$$yn=m=\log n$$
$$y=\frac{\log n}{n}$$
$$\log x=\frac{\log n}{n}$$
$$x=e^{\frac{\log n}{n}}=\left(e^{\log n}\right)^{\frac1n}=n^{\frac1n}=\sqrt[n]{n}$$
For example, if $n=5$ then we have $x=\sqrt[5]{5}$, like you found.
If $x$ or $n$ is a complex number, or if $n<e^{-1}$, then the argument
$$yne^{yn}=me^m\iff yn=m$$
fails, because $y=xe^x$ ceases to be injective. In that case, you may need to use the Lambert W function. $x=\sqrt[n]{n}$ will remain a solution, but it will not be the only solution.
$\endgroup$ $\begingroup$Hint Let $y=x^5$, then you have $(y^{1/5})^y=5\iff y^{y/5}=5\iff y^y=5^5$
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