Proving 2^n < n without Equality: Why Ask?

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The discussion centers on proving the inequality 2^n < n! for all n ≥ 4. The original poster successfully proved 2^n ≤ n! using induction but struggled to demonstrate the strict inequality without equality. Participants clarify that showing 2^n < n! implies 2^n ≤ n! is valid, and the textbook's request for the weaker condition may be a typo. They confirm that for n = 4, 2^4 is indeed less than 4!, reinforcing the inequality. Overall, the consensus is that the condition n ≥ 4 is accurate, while the equality aspect may be an error.
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In my maths textbook it asks to prove 2^n<=n! for all n>=4

I could prove it no problems using induction but could show 2^n<n! without the equality inequality.

My question is why would the textbook ask for a weaker condition? Is it a misprint?

If I can show < then it automatically implies <= holds as well dosen't it?
 
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If I can show < then it automatically implies <= holds as well dosen't it?
Yes, if x is strictly less than y, then x is certainly less than or equal to y. With regards to your former question, it might just be a typo- either way, you can solve the problem using induction.
 
It can't be a missprint. See what happens with n=4.

Daniel.

PS. Apparently, it can be a missprint.
 
Last edited:
dextercioby said:
It can't be a missprint. See what happens with n=4.

Daniel.

2^4=16 , 4!= 24, so 2^4<4!
 
I was only asking the relationship between 2^n and n!

Looking at the graphs for 2^n and n! it seems that no where is 2^n=n!

n>=4 is definitely correct.
 
Look at what are the factors of 2^n and the factors of n!...
 
There are two "=" signs in question. The first one is definitely a misprint but n>=4 is correct.
 

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