I'm having difficulty proving $n! > n^2$ for $n \ge 4$ I have previously solved a similar problem but it is $n! > 2^n$. Now I don't know how to solve this. I have only come as far as solving for the base case. Thank you so much for helping a struggling student out.
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1See also this post and other questions linked there. – Martin Sleziak Oct 27 '16 at 07:14
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So the problem is $$n!>n^2 \space \forall n\ge4$$
We'll try to tackle it using induction:
Base case: $$4!>16$$
Hypothesis: $$n!>n^2$$
Induction: $$(n+1)!>(n+1)^2 \Rightarrow (n+1)(n)!\gt(n+1)^2 \Rightarrow n!\gt\frac{(n+1)^2}{(n+1)} \Rightarrow \bbox[border:1px solid red]{n!\gt n+1}$$
The induction is proved.
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1$(n+1)!$ is $(n+1)(n)(n-1)...(2)(1)$ right? Therefore it can also be seen, by "taking away" $(n+1)$ from the factorial, as simply $(n+1)$ times my old factorial $n!$ – Francesco Carzaniga Oct 27 '16 at 07:19
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Hint: Try to justify \begin{align} n(n^2-2n+1) \ge n^2. \end{align} for $n\geq 4$.
Jacky Chong
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