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Mathematical induction proof?

what is the most simplest method to solve this question by using mathematical induction to prove that \(n !>2^n-1 \) and integers \(n>5\)?

Note by Syed Hissaan
3 weeks, 4 days ago

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@Calvin Lin @Chung Kevin , any proof for that one ! Syed Hissaan · 3 weeks, 4 days ago

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@Syed Hissaan What have you tried? What have you done?

Seems to me like a simple induction proof starting with base case of \( n = 6 \). Calvin Lin Staff · 3 weeks, 4 days ago

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@Calvin Lin is it valid in this type of proof to place any value of the k < 6 on one of the side to setup our equations)# for n=k e.g : \( (k+1)!> 2^k .2 -1 +2^k -k \) can we take k<4 * or *k=2on the R.H.S to proof our question Syed Hissaan · 3 weeks, 3 days ago

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@Syed Hissaan No, because the statement is not true for \( k \leq 4 \). That is why you have to start the base case in the appropriate domain of \( n > 5 \) (or whatever they tell you to). Calvin Lin Staff · 3 weeks, 3 days ago

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@Calvin Lin ok thanks ! Syed Hissaan · 3 weeks, 3 days ago

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