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Inductive case: Let \(k \ge 5\) be an arbitrary integer. Assume, for induction, that \(P(k)\) is true. That is, assume \(k^2 \lt 2^k\text{.}\) We will prove that \(P(k+1)\) is true, i.e., \((k+1)^2 \lt 2^{k+1}\text{.}\) To prove such an inequality, start with the left-hand side and work towards the right-hand side:

\begin{align*} (k+1)^2 \amp = k^2 + 2k + 1 \amp\\ \amp \lt 2^k + 2k + 1 \amp \ldots\text{by the inductive hypothesis.}\\ \amp \lt 2^k + 2^k \amp \ldots\text{ since } 2k + 1 \lt 2^k \text{ for }k \ge 5.\\ \amp = 2^{k+1}. \amp \end{align*}

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