If $x$ is a real number, then solve for $x$ wherever the below equation is defined: $$x^{x-1} - (x-1)^x = 1$$
I viewed the answer to the question: $x^{x-1} = (x-1)^x $; but those methods won't work here because of the $1$ on RHS. Also by using a graph plotter, I found that the only solutions for $x \geq 1$ are $x = 1,2$ or $3$. How to prove this, and how to find the solutions for $x < 1$?
Partial solution: A standard result here is that $x^y > y^x$ if $e \le x < y$. Therefore there are no solutions if $x > e+1$.