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I will prove the false statement, that [tex]n\geq a\Rightarrow n!\geq a^n, n\in \mathbb{N}-\left \{ 0 \right \}[/tex] with induction

For [tex]n=1[/tex] [tex]1\geq a\Rightarrow 1!\geq a^1\Rightarrow 1 \geq a[/tex] which is true.

Suppose that [tex]n\geq a\Rightarrow n!\geq a^n, n\in \mathbb{N}-\left \{ 0 \right \}[/tex]

Then,

[tex]n\geq a\Rightarrow (n+1)!\geq nn!\geq aa^n=a^{n+1}[/tex] which yields that [tex]n+1\geq a\Rightarrow(n+1)!\geq a^{n+1}[/tex]

Therefore, [tex]n\geq a\Rightarrow n!\geq a^n[/tex]

But for [tex]n=3,a=2[/tex] using the inequality we just proved [tex]3\geq 2\Rightarrow3!\geq 2^3\Leftrightarrow 6\geq 8[/tex] Impossible!!. Where is my mistake?

[EDIT] Don't bother answering. I have highlighted the mistake I made that rendered the inequality invalid for a greater than 1

For [tex]n=1[/tex] [tex]1\geq a\Rightarrow 1!\geq a^1\Rightarrow 1 \geq a[/tex] which is true.

Suppose that [tex]n\geq a\Rightarrow n!\geq a^n, n\in \mathbb{N}-\left \{ 0 \right \}[/tex]

Then,

[tex]n\geq a\Rightarrow (n+1)!\geq nn!\geq aa^n=a^{n+1}[/tex] which yields that [tex]n+1\geq a\Rightarrow(n+1)!\geq a^{n+1}[/tex]

Therefore, [tex]n\geq a\Rightarrow n!\geq a^n[/tex]

But for [tex]n=3,a=2[/tex] using the inequality we just proved [tex]3\geq 2\Rightarrow3!\geq 2^3\Leftrightarrow 6\geq 8[/tex] Impossible!!. Where is my mistake?

[EDIT] Don't bother answering. I have highlighted the mistake I made that rendered the inequality invalid for a greater than 1

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