getting phi of a large number

hanelliot

phi(20^100)
= phi(4^100 * 5^100)
= phi(2^200 * 5^100)
= (2^200 - 2^199)(5^100 - 5^99)
= 2^199(2-1) * 5^99(5-1)
= 2^199 * 5^99 * 4
= 2^201 * 5^99.

I don't understand line 4-7. Can anyone explain?

CRGreathouse

phi(p^a) = (p-1)(p^(a-1)) for p prime

matticus

also since phi is multiplicative phi(a*b) = phi(a)*phi(b) when gcd(a,b)=1