Is \( n! > 2^n \) for \( n \ge 4 \)?

[karush]

$\tiny{2.3.3}$
Prove $n! > 2^n$ for $n\ge4$
ok well the only thing I know to do is just try some numbers

$n=4, \quad 4\cdot3\cdot2\cdot1=24 \quad 2^4=32 \quad \therefore 32\ge24$
$n=5, \quad 5\cdot4\cdot3\cdot2\cdot1=120 \quad 2^5=32 \quad \therefore 120\ge32$

not sure just what the proof would be and looks like it must just be intergers

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[topsquark]

$\tiny{2.3.3}$
Prove $n! > 2^n$ for $n\ge4$
ok well the only thing I know to do is just try some numbers

$n=4, \quad 4\cdot3\cdot2\cdot1=24 \quad 2^4=32 \quad \therefore 32\ge24$
$n=5, \quad 5\cdot4\cdot3\cdot2\cdot1=120 \quad 2^5=32 \quad \therefore 120\ge32$

not sure just what the proof would be and looks like it must just be intergers

uld be

Are you allowed to use induction? Let k be the lowest integer such that \(\displaystyle k! > 2^k\). Then show that \(\displaystyle (k + 1)! > 2^{k + 1}\)

-Dan

 

[karush]

Re: aa2.3.3 Prove n! > 2^n for n\ge4

Are you allowed to use induction? Let k be the lowest integer such that \(\displaystyle k! > 2^k\). Then show that \(\displaystyle (k + 1)! > 2^{k + 1}\)

-Dan

I thot that is what i did?

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[topsquark]

You showed a couple of examples which show a definite trend but didn't give an actual proof. This is a simple induction proof and basically copies what you have shown.

-Dan

 

[Olinguito]

Since $n\ge4=2^2$,
$$n!\ =\ \underbrace{n}_{\ge2^2} \cdot \underbrace{(n-1)}_{>2} \cdot \cdots \cdot \underbrace{3}_{>2} \cdot \underbrace{2}_{\ge2} \cdot 1\ >\ 2^n.$$