By using mathematical induction prove that (n+1)! > 2^(n+1) for n, where n is a positive integer greater than or equal to 4
Prove (n+1)!>2(n+1)(n+1)! > 2^{(n+1)}(n+1)!>2(n+1) :
Base case: n=4n = 4n=4
(4+1)!=5!=120>2(4+1)=25=32(4+1)! = 5! = 120 > 2^{(4+1)} = 2^5 = 32(4+1)!=5!=120>2(4+1)=25=32 -> true
suppose for some n: n!>2nn! > 2^nn!>2n , let's prove that (n+1)!>2(n+1)(n+1)! > 2^{(n+1)}(n+1)!>2(n+1)
(n+1)!=n!∗(n+1)(n+1)! = n! * (n+1)(n+1)!=n!∗(n+1)
2(n+1)=2∗2n2^{(n+1)} = 2 * 2^n2(n+1)=2∗2n
using assumption that n!>2nn! > 2^nn!>2n it is obvious that n!∗(n+1)>2∗2nn! * (n+1) > 2 * 2^nn!∗(n+1)>2∗2n , since n+1>2n+1 > 2n+1>2 for n≥4n \ge 4n≥4 .
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