Base case, n=1, (a/b)^1 = a^1/b^1
Assumption: (a/b)^n = a^n/b^n
then to prove (a/b)^(n+1) = a^(n+1)/b^(n+1)
LHS = (a/b)^(n+1)
=> (a/b)*(a/b)^(n)
=> (a/b)*(a^n/b^n) , Assumption
=> (a*a^n)/(b*b^n)
=> a^(n+1)/b^(n+1) = RHS
Hence by mathetical induction (a/b)^n = a^n/b^n. Proved.