Answer
See below.
Work Step by Step
Proofs using mathematical induction consists of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we prove that then the statement also holds for $n + 1$.
Hence here:
1) For $n=1: (\frac{a}{b})^1=\frac{a^1}{b^1}$, which is obviously true.
2) Assume for $n=k: (\frac{a}{b})^k=\frac{a^k}{b^k}$. Then for $n=k+1:(\frac{a}{b})^{k+1}=\frac{a^{k+1}}{b^{k+1}}$.
By the two known statements:
$(\frac{a}{b})^{k+1}=(\frac{a}{b})^{k}\cdot(\frac{a}{b})^{1}=\frac{a^k}{b^k}\frac{a^1}{b^1}=\frac{a^{k+1}}{b^{k+1}}.$ Thus we proved what we wanted to.