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How do I do this?

I know how to do the base case but I can't figure out how to do the next steps.

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closed as off-topic by Xander Henderson, John Omielan, max_zorn, Dbchatto67, Leucippus Apr 19 at 5:48

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Hint: We can write $$2^{n-2}\cdot n<4\cdot n!$$ by cross multiplication and this is $$2^{n-2}<(n-1)!$$, which is easier to prove. and we have to prove that $$2^{n-1}<n!$$ if $$2^{n-2}<(n-1)!$$ multiplying this inequality by $2$ we get $$2^{n-1}<2(n-1)!$$ and $$2(n-1)!<n!$$ for $$n>2$$

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