Answer
See below.
Work Step by Step
Proofs using mathematical induction consist 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 the statement also holds for $n + 1$.
Hence, here we have:
1) For $n=1: 3^{2(1)}-1=8$ is divisible by $8$.
2) Assume for $n=k: 3^{2k}-1$ is divisible by $8$
Then for $n=k+1$: $3^{2(k+1)}-1=9\cdot3^{2k}-1=9\cdot3^{2k}-9+8=9(3^{2k}-1)+8$
Notice that $3^{2k}-1$ is divisible by $8$ by the induction hypothesis above. Multiplying this number by $9$ will not change its divisibility by $8$. Similarly, adding $8$ will mean that it is still divisible by 8.
Thus, we proved what we had to.