The question:
Prove that for all positive integers $n \geq 4$, $2^{n} < n!$
The base case is fine, but how do you proceed with inductive step?
$(2^{k} < k!) \implies ( 2^{k+1} < (k+1)!)$
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For a good look at induction proofs, see this great answer – Ross Millikan Jan 31 '19 at 00:30
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1You've written the inductive step oddly. To keep the logic straight in your head, it might help to write correctly: $$(2^{k} < k!) \implies ( 2^{k+1} < (k+1)!)$$ or, in words, $$2^{k} < k! ,,\text{implies},, 2^{k+1} < (k+1)!$$ – Lee Mosher Jan 31 '19 at 00:32
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Yes you're right. Sloppy on my end. Will change. – Mark Jan 31 '19 at 00:37
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Intuitively, to go from $n!$ to $(n+1)!$ you multiply by $n+1$, which is greater than $2$.
Ross Millikan
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