Proof by induction: n^3 < n for n >=6

[1MileCrash]

Homework Statement

Show that n^3 < n! for all n >= 6.

Homework Equations

The Attempt at a Solution

We see that for the base case of n = 6, the claim holds.

Suppose that k^3 < k! for some natural number k >= 6.

Consider that:
(k+1)^3
= k^3 + 3k^2 + 3k + 1
< k! + 3k^2 + 3k + 1 [By induction hypothesis]

What's a neat way to finish this? I'm a bit rusty, apparently.[/B]

 

[Dick]

Homework Statement

Show that n^3 < n! for all n >= 6.

Homework Equations

The Attempt at a Solution

We see that for the base case of n = 6, the claim holds.

Suppose that k^3 < k! for some natural number k >= 6.

Consider that:
(k+1)^3
= k^3 + 3k^2 + 3k + 1
< k! + 3k^2 + 3k + 1 [By induction hypothesis]

What's a neat way to finish this? I'm a bit rusty, apparently.[/B]

Factor a $k^3$ out of $k^3+3k^2+3k+1$. Then use $k^3 \le k!$ and $(k+1)!=k!(k+1)$.

 

[1MileCrash]

Alright. And then, what remains after being factored has its largest value at k=6, and its value is smaller than any (k+1), and so I may write < k!(k+1), completing the induction.

 

[Dick]

Alright. And then, what remains after being factored has its largest value at k=6, and its value is smaller than any (k+1), and so I may write < k!(k+1), completing the induction.

Sounds ok.