Inductive case: Let \(k\) be an arbitrary natural number. Assume, for induction, that \(P(k)\) is true. That is, \(6^k - 1\) is a multiple of \(5\text{.}\) Then \(6^k - 1 = 5j\) for some integer \(j\text{.}\) This means that \(6^k = 5j + 1\text{.}\) Multiply both sides by \(6\text{:}\)