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 $n=1: (ab)^1=a^1b^1$.
2) Assume for $n=k: (ab)^k=a^kb^k$. Then for $n=k+1:(ab)^{k+1}=a^{k+1}b^{k+1}$.
By the two known statements:
$(ab)^{k+1}=(ab)^k\cdot(ab)=(a^kb^k)(a^1b^1)=a^{k+1}b^{k+1}$Thus we proved what we wanted to.