Suppose we have no idea that $x^{1/x}=\exp \left( \frac{1}{x} \ln x \right)$ or don't know anything about exponents and logarithms.
But we can certainly compute roots, for example by this method.
Then we can solve this equation the following way:
$$x^{1/x}=y^{1/y}$$
Denote $\sqrt[y]{y}=z$.
$$x=z^x=z^{z^x}=z^{z^{z^{z^x}}}==z^{z^{z^{z^{\cdots}}}}$$
Really, we just do the following backwards recursion:
$$x_0=1$$
$$x_{k+1}=z^{x_k}$$
We know that the function $f(x)=x^{1/x}$ reaches its maximum at $x=e$. So let's try some numbers $y>e$. I take exactly $k=19$ steps.
$$y=3$$
$$x_{19}=2.41 \dots$$
Numerical solution by WolframAlpha:
$$x(3)=2.4780527 \dots$$
So the convergence is slow - only one correct digit.
$$y=4$$
$$x_{19}=1.9994 \dots$$
$$x=2$$
Here we have three correct digits.
$$y=7$$
$$x_{19}=1.53014008 \dots$$
Numerical solution
$$x=1.53014012 \dots$$
And we already have 7 correct digits.
Why does this algorithm converge better for larger $y$? Why does it converge extremely slowly for $y$ close to $e$?
Edit
As for the irrational exponentiation - it can also be defined without the exponent, since we only use the rational approximations to the roots anyway.
HINT.- $f(x)=x^{\frac 1x}$ has domain $x\gt o$. Assuming the following facts:
►$\lim_{x\to \infty}f(x)=1\\►f(1)=1\\►f(0)=0\space\text {(prolongement by continuity or limit)}\\►\max f(x)=f(e)\approx 1.444668$.
It follows:
(1) $f$ is injective for $0<x\le 1$ so $x^{\frac 1x}=y^{\frac 1y}\iff x=y$
(2) if $1<x<e$ then $x^{\frac 1x}=y^{\frac 1y}$ has a unique solution $y\ne x$; for $x$ near to $1$, the solution $y$ is quite big; for $x$ near to $e$, the solution $y$ is near to $x$
(3) if $x=e$ then $e^{\frac 1e}=y^{\frac 1y}$ has only the solution $y=e$
I hope these remarks can help the OP Yuriy S.