$2^n<n!$ for some $n>3\in\mathbb Z^+$
Basic step is true with $n=4$.
So induction step assume $2^n<n!$ for some $n>3\in\mathbb Z^+$ is true.
Will show that $2^{n+1}<(n+1)!$ for some $n>3\in\mathbb Z^+$.
Consider $2^{n+1}=(2)2^n$
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15 seconds of searching gave at least four earlier duplicates of this question. Downvotes to established answerers for not caring about the site hygiene at all. – Jyrki Lahtonen Jun 13 '18 at 04:50
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If $n! >2^n$ for some $n>3$, then $(n+1)!=n!(n+1)>2^n(n+1) > 2^n \cdot 2 =2^{n+1}$, since $n+1>2$.
Fred
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+1 to offset [what looks to me like] an unreasonable downvote – Benjamin Dickman Jun 13 '18 at 04:44
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You're welcome; I see that we posted answers in the same spirit and were both downvoted. Maybe because the question was closed as a duplicate? [Considering the timing, I would view this as an unjustified reason to downvote. Nevertheless...] – Benjamin Dickman Jun 13 '18 at 04:50
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1@BenjaminDickman I downvoted you both. You have been around the site for long enough to at least suspect that this question has been asked before. Yet, you apparently thought it more important to first compile an answer. I don't want such behavior to be rewarded. I wish there were more positive ways of causing a change. Too many users don't care at all. If you actually do, then I apologize. Whenever an answer is posted within ten minutes Occam's razor tells me that... – Jyrki Lahtonen Jun 13 '18 at 04:56
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It seems that your argument ended rather abruptly.
Under the supposition that $n>3$, you would have that:
$$2^{n+1} = 2^n \cdot 2 < n! \cdot (n+1) = (n+1)!$$
where the inequality follows since $2^n < n!$ using the inductive hypothesis and $2 < n+1$ since we are in the case of $n > 3$. This proves $P(n) \implies P(n+1)$ for the suitable proposition $P$, which combines with a base case of $n=4$ to show that the proposition holds for all integers $n > 3$ by the principle of mathematical induction. QED
Benjamin Dickman
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