Why??

easy

0=n-n

a^0=a^(n-n)

a^0=a^n * a^(-n)=a^n/a^n=1

for any 'n' ....

oh....it is thus defined...

o⁰is inderteminate
beacause
same logic given by prateek
o⁰ can be written as
0^1-1
i.e.
0/0
and we all know that 0/0 is indeterminant.
thus it is not defined.

There is a TiiT Classic on 0⁰...
( http://targetiit.com/iit-jee-forum/posts/10th-january-2009-1565.html )

Hmmm , pretty good and simple doubt posted here , ...... , nice to see .

As far as ( 0 ) ⁰ , is concerned , I think the answer can either be 1 or 0 depending on the conditions .

One should probably note that ( 0 ) ⁰ is actually the single element in the intersection of the sets ,

" 0 ⁿ " and " n ⁰ " , where " n " is any number .