Proof by induction that $n^2 < n!$

Question

Prove $n^2 < n!$.

This is what I have gotten so far

basis step: $p(4)$ is true Inductive Hypothesis assume $p(k)$ true for $k \ge 4$

Inductive Step $p(k+1)$ : $(k+1)^2 < (k+1)!$

$$(k+1)^2 =k^2 + 2k + 1 < k! + 2k +1$$

Can someone please explain the last step this is from text, I need help understanding this, forgive me for the formatting error Im still learning

See http://math.stackexchange.com/questions/1140396/prove-by-induction-that-n2n and http://math.stackexchange.com/questions/1462423/proving-n-n2-by-mathematical-induction — Martin Sleziak, Oct 25 '15 at 07:02

score 4 · Accepted Answer · answered Sep 20 '14 at 18:18

4

Inductive Step:

Assume the case for $n$ is true, then for $n \geq 4$ $$(n + 1)^2 = n^2 + 2n + 1 < n! + 2n + 1 < n! + n^2 \leq n! + n!n = n!(n+1) = (n+1)!.$$

answered Sep 20 '14 at 18:18

IAmNoOne

1

@MauroALLEGRANZA, I misread! I fix rright now. This is what I get for being lazy. – IAmNoOne Sep 20 '14 at 18:20

score 1 · Answer 2 · answered Sep 20 '14 at 18:19

$$n^n \geq n!$$

Proof: Let $n\in\mathbb{N}$. Then $$ n^n = n\cdot n\cdot n\cdot...\cdot n$$ where as $$ n! = n\cdot(n-1)\cdot (n-2)\cdot...\cdot1$$

For each term in the product you can compare $$n = n $$ $$n > n-1 $$ $$n > n-2 $$ and so on. Thus $n^n \geq n!$

2 Answers2