Prove the statement by the Principle of Mathematical Induction: $n^{2} < 2^{n}$ for all natural numbers $n \geq 5$.
(N/A) Let $P(n): n^{2} < 2^{n}$ for $n \geq 5, n \in \mathbb{N}$.
Step $1$: For $n = 5$,$P(5): 5^{2} < 2^{5} \implies 25 < 32$,which is true.
Step $2$: Assume $P(k)$ is true for some $k \geq 5$,i.e.,$k^{2} < 2^{k}$.
We need to show $P(k+1): (k+1)^{2} < 2^{k+1}$ is true.
Consider $(k+1)^{2} = k^{2} + 2k + 1$.
Since $k^{2} < 2^{k}$,we have $(k+1)^{2} < 2^{k} + 2k + 1$.
For $k \geq 5$,it can be shown that $2k + 1 < k^{2} < 2^{k}$.
Thus,$(k+1)^{2} < 2^{k} + 2^{k} = 2 \cdot 2^{k} = 2^{k+1}$.
Therefore,$P(k+1)$ is true whenever $P(k)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n \geq 5$.
Step $1$: For $n = 5$,$P(5): 5^{2} < 2^{5} \implies 25 < 32$,which is true.
Step $2$: Assume $P(k)$ is true for some $k \geq 5$,i.e.,$k^{2} < 2^{k}$.
We need to show $P(k+1): (k+1)^{2} < 2^{k+1}$ is true.
Consider $(k+1)^{2} = k^{2} + 2k + 1$.
Since $k^{2} < 2^{k}$,we have $(k+1)^{2} < 2^{k} + 2k + 1$.
For $k \geq 5$,it can be shown that $2k + 1 < k^{2} < 2^{k}$.
Thus,$(k+1)^{2} < 2^{k} + 2^{k} = 2 \cdot 2^{k} = 2^{k+1}$.
Therefore,$P(k+1)$ is true whenever $P(k)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n \geq 5$.
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