y = Z^N

This little script is for calculating Z^N, where Z and N are numeric inputs. However, I will not use any power or root functions, and will base my script only on addition, substraction, multiply, division and mod. It's also as efficient as possible, and is based on the logic equivalences [a^(b+1) = a^b * b] and [a^(2*b) = (a^2)^b]. Check the source for the code and it's valid proof.

Z:
N:
y: