Is $(0,0)$ a solution to $x^y-y^x=0$? I am trying to determine if $(0,0)$ a solution to $x^y-y^x=0$.  My hunch is that it is undefined since $0^0$ is an indeterminate form.  To attempt to prove this, I have tried the usual "different paths give different limits" trick with
$(x,ax^n)\rightarrow(0,0)$
$(x,\sin(x))\rightarrow(0,0)$
$(x,e^x-1)\rightarrow(0,0)$
None of the above accomplished my goal.  
I did find this older post ($x^y = y^x$ for integers $x$ and $y$) which included an answer by "Yuval Filmus" which states $y=x=0$ is $\it{trivially}$ a solution.  He accomplished this by defining $0^0$ to some value and moves on.  I would like to see something more rigorous, if it exists.
Any hints on how to proceed?
Edited for wording.
Edit:  Stefan Smith confirms my suspicion that the comment of Yuval Filmus cannot be made rigorous.  

This might sound unsatisfying, but if you define $0^0$ to be any finite number, then $(0,0)$ is a solution, otherwise it isn't. I personally like $0^0=1$, but some people disagree. This site has a lot to say about the issue of $0^0$, and you can search for that. It is ultimately a matter of definition. 
