The issue of 0^0 still intrigues me.

0^n = 0 for all n>0
0^n is undefined for n<0
But from your argument you state that 0^0 = 1.

So while f(x) = x^n is usually a continuous function, the results above imply that this is only true if x does not equal 0.

Is this a simple explanation as to why 0^0 is left undefined since a base of 0 breaks the rules of continuity anyway?

I am very hazy on my limits but it seems strange to contemplate that lim n approaches 0 of 0^n = 1 when the nth root of 0 is 0.