For a natural number n,(n!)^2 n^n is true i | vedclass

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Question

(D) We test the inequality $(n!)^2 > n^n$ for natural numbers $n \geq 1$: For $n=1$: $(1!)^2 = 1$ and $1^1 = 1$. Since $1 > 1$ is false,the inequality

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$n \geq 3$

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(D) We test the inequality $(n!)^2 > n^n$ for natural numbers $n \geq 1$: For $n=1$: $(1!)^2 = 1$ and $1^1 = 1$. Since $1 > 1$ is false,the inequality does not hold. For $n=2$: $(2!)^2 = 4$ and $2^2 = 4$. Since $4 > 4$ is false,the inequality does not hold. For $n=3$: $(3!)^2 = 36$ and $3^3 = 27$. Since $36 > 27$ is true,the inequality holds. For $n=4$: $(4!)^2 = 576$ and $4^4 = 256$. Since $576 > 256$ is true,the inequality holds. Thus,the inequality $(n!)^2 > n^n$ is true for all $n \geq 3$.

For a natural number $n$,$(n!)^2 > n^n$ is true if:

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