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

1. Mar 8, 2015

### 1MileCrash

1. The problem statement, all variables and given/known data
Show that n^3 < n! for all n >= 6.

2. Relevant equations

3. 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.

2. Mar 8, 2015

### Dick

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

3. Mar 9, 2015

### 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.

4. Mar 9, 2015

Sounds ok.