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Prove: $n! > 2^n$ for all $n \geq 4$

I already proved the base case: $24 > 16$.

Then I assume this holds for $n=k$, and start proving it for $n=k+1$:

$(k+1)k! > 2^{k+1}$

$(k+1)k! > 2^{k}(2^1)$

After this I'm not sure how to proceed. Any help is appreciated, thanks in advance.

N. F. Taussig
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user360431
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    You need to prove that $k!>2^k$ implies $(k+1)!>2^{k+1}$. You cannot do that by first assuming that $(k+1)!>2^{k+1}$. – Angina Seng Oct 08 '18 at 06:27

3 Answers3

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Well, notice that

$$ (k+1)! = (k+1) k! \geq (k+1) 2^k = k 2^k + 2^k > 2^k+2^k=2\cdot2^k= 2^{k+1} $$

James
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The induction hypothesis is that $k!>2^k$ for some $k\geq4$. Then $(k+1)! =k!(k+1) > 2^k(k+1)$. But $2^k(k+1)>2^k2$, since $k\geq 4$, and so $k(k+1)!>2^{k+1}$.

Wuestenfux
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Step :

$2(n!) >2 \cdot 2^n=2^{n+1}$(hypothesis).

For $n \ge 4:$

$(n+1)!= (n+1)n! > 2n! > 2^{n+1}.$

Peter Szilas
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