markr is right
0^0 is
indeterminate.
As a second source (Dr Math) alluded, any number n^0, except 0, can be interoperated as a logarithm (mathematical transcendentalism)
log(basen)(n^0)=0*1=0
and since n^0=1
for any number except 0
Consider logarithms
log(basen)X as n -> 0=infinity
then log(base0)= 0*infinity
which is as indeterminate as 0/0.
So 0^0, like 0/0, is undefined/indeterminate
Rap