n² < 2ⁿ for all natural numbers n ≥ 5.

Given; P(n) is n² < 2ⁿ ; for n≥5

Let P(k) = k² < 2^k is true;

⇒ P(k+1) = (k+1)²

= k² + 2k + 1

2^k+1 = 2(2^k) > 2k²

∵ n² > 2n + 1 for n ≥3

∴ k² + 2k + 1 < 2k²

⇒ (k+1)² < 2^(k+1)

⇒ P(k+1) is true when P(k) is true.

∴ By Mathematical Induction P(n) = n² < 2ⁿ is true for all natural numbers n ≥ 5.