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I have to prove by induction that $n! > n^3.$ This how far I've reached:

Induction Hypothesis: $k! > k^3$.

Induction step: Basically for Inductive step, I have to show that, $(k+1)! > (k+1)^3$.

L.S = $(k+1)! = k!(k+1)$

Where do I go from here? Any clues are appreciated. Thank you.

amWhy
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Muttaki
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    Welcome to Mathematics Stack Exchange. To get from $k!>k^3$ to $(k+1)!>(k+1)^3$, you multiply the left side by $k+1$ and the right side by $(1+\frac1k)^3$; can you show $k+1>(1+\frac1k)^3$? – J. W. Tanner Jan 30 '20 at 13:34
  • @J.W.Tanner Brooo thank you so much. Your idea actually helped! God bless you. – Muttaki Jan 30 '20 at 13:47

2 Answers2

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The induction from $k-1$ to $k$ is a little simpler.

$$k!=k(k-1)!>k(k-1)^3=k^3\frac{(k-1)^3}{k^2}=k^3\left(k-3+\frac{3k-1}{k^2}\right).$$

The last factor certainly exceeds $1$ as of $k=4$.

For a non-inductive proof,

$$n!=2\cdot3\cdots(n-2)(n-1)n>n^3$$ follows from

$$2(n-2)(n-1)>n^2$$

or $$n^2-6n+4>0$$

which is true for $n\ge6$.

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For $k>5, $ $k>1+1+1>\frac3k+\frac3{k^2}+\frac1{k^3}$, so $k+1>(1+\frac1k)^3.$

Therefore, $k!>k^3\implies (k+1)!>(k+1)^3$.

J. W. Tanner
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