Proof.
Let \(P(n)\) be the statement \(\log(a^n) = n \log(a)\text{.}\) The base case, \(P(2)\) is true, because \(\log(a^2) = \log(a\cdot a) = \log(a) + \log(a) = 2\log(a)\text{,}\) by the product rule for logarithms. Now assume, for induction, that \(P(k)\) is true. That is, \(\log(a^k) = k\log(a)\text{.}\) Consider \(\log(a^{k+1})\text{.}\) We have
\begin{equation*}
\log(a^{k+1}) = \log(a^k\cdot a) = \log(a^k) + \log(a) = k\log(a) + \log(a)\text{,}
\end{equation*}
with the last equality due to the inductive hypothesis. But this simplifies to \((k+1) \log(a)\text{,}\) establishing \(P(k+1)\text{.}\) Therefore by the principle of mathematical induction, \(P(n)\) is true for all \(n \ge 2\text{.}\)