It can be solved graphically using the intersection point of $y= \log_ x (n)$ and $y=x$.
For $n=2$ the intuitive solution is easy its $x=2$ only but how to solve it more generally for $n \in \mathbb{R}$
Reframing the question:
How can I solve $x^x=n$,for some $n \in \mathbb{R}$ algebraically?
Thank you!
Thought this looked fairly simple but after a few lines of working realised I've never had to approach anything of that form before.
In this case, it is easy to inspect that $x = 2$ is the only solution, but for a more general $x ^ x = n$, for some $n \in \mathbb{R}$, the solution is a bit harder to find.
It turns out there is no standard function (polynomial, trigonometric, logarithmic) which can solve this, so we resort to numerical methods.
The solution can be found using the Newton-Raphson method, which recursively converges on a solution to your equation (though if the initial input does not meet certain conditions then this may not work). This will obviously never be perfectly accurate, so we can represent the true value using the Lambert W function In short, there is no "exact" quick, easy and general solution, but you may want to look further into the Lambert W function and the Newton method of approximation.