# Proof by induction that $n^2 < n!$ [duplicate]

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

• its 2 sorry not n Sep 20, 2014 at 18:12
• In the last step, you are using the inductive hypothesis. Sep 20, 2014 at 18:22
• Can you explain how is that true < (k+1)! ? Sep 20, 2014 at 18:25
• I explained that in my answer below... Sep 20, 2014 at 18:25
• Oct 25, 2015 at 7:02

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