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

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Prove $(n+1)! > 2^{(n+1)}$ :

Base case: $n = 4$

$(4+1)! = 5! = 120 > 2^{(4+1)} = 2^5 = 32$ -> true

suppose for some n: $n! > 2^n$ , let's prove that $(n+1)! > 2^{(n+1)}$

$(n+1)! = n! * (n+1)$

$2^{(n+1)} = 2 * 2^n$

using assumption that $n! > 2^n$ it is obvious that $n! * (n+1) > 2 * 2^n$ , since $n+1 > 2$ for $n \ge 4$ .