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I am stuck with the question below,

Prove by mathematical induction that $n for $n>2$.

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    Akito was missing the $'s in his LaTeX.2011-10-15

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First, for $n=3$ you have $3< 3!=6 $. Suppose that for some $k$ it holds that $k then $ (k+1)! = (k+1)k!>(k+1)k\geq k+1 $ since $k\geq 3$. Could you please tell which step is unclear to you in this proof? By elaborating on it maybe we can learn how to use induction.

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    The thing written above is the correct answer. I don't know where all these comments come from.2012-11-27
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If this is homework and the professor specifically said to use induction, then disregard this answer, I suppose. Otherwise, the statement can be proven directly without induction.

Given any $n \geq 3$, we can write $n! = n(n-1)!$ and be confident that $n-1 \geq 2$ (so we aren't making inappropriate use of $0!$). From this expression, it is clear that $n! > n$, since $n!$ is equal to $n$ times some number strictly greater than 1.

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    @HenningMakholm: (n-1)!>1 follows from (n-1)!>n-1 but not vice versa. I mean that he deduces like this: n! = n(n-1)!>n only using (n-1)!>1 rather than more 'stronger' statement (n-1)!>n-1. Though, that's hard for me to understand what is simpler here. Proofs here are quite indistinguishable2011-10-15