what is the most simplest method to solve this question by using mathematical induction to prove that \(n !>2^n-1 \) and integers \(n>5\)?
11 months, 2 weeks ago
@Calvin Lin @Chung Kevin , any proof for that one !
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What have you tried? What have you done?
Seems to me like a simple induction proof starting with base case of \( n = 6 \).
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
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).
ok thanks !