Question 168709
See this --> http://mathforum.org/dr.math/faq/faq.number.to.0power.html

x^0 where x is anything except 0 is defined as 1

You can see that {{{x^0 = x^(y-y)}}}
{{{x^(y-y) = (x^y)/(x^y)= 1}}} for all x!=0
this works for all x except x=0. If x=0 then term (0^y) in the denominator is undefined