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 $n=1$: $a^1-b^1=a-b=(a-b)1$, thus $(a-b)$ is a factor of it.
2) Assume for $n=k$: $a^k-b^k$ has $(a-b)$ as a factor. Then for $n=k+1$:
$a^{k+1}-b^{k+1}=a(a^k-b^k)+b^k(a-b)$. $(a-b)$ divides both elements in the sum, thus it will also divide their sum.
Thus we proved what we wanted to.