# Prove by induction $n! > n^2$

I am trying to prove the inequality in the title for $n\geq 4$; however, I am stuck on the induction step! Any help would be appreciated.

For $n\ge 4$, prove that $n! > n^2$.

Base Case: $n=4$, LHS $4! = 24$, RHS = $4^2 = 16$

$24>16$ : True

Induction Hypothesis: Assume True for $n=k$.

$k! > k^2$

Induction Step: Should be True for $n=k+1$

$(k+1)! > (k+1)^2$

$(k+1) . (k)! > (k^2 + 2k + 1)$

However, here is where I get stuck.

• $k! > k^2 \Rightarrow (k+1)k! > (k+1)k^2$. Now, can you show that for $k \geq 4$ we have $$(k+1)k^2 > (k+1)^2$$ That is $$k^2 > k + 1$$ – Simon S Apr 27 '15 at 12:15
• @GeorgeS wrong question... – Clement C. Apr 27 '15 at 12:16
• Yes, I just realised. Sorry that was an accident. – user230715 Apr 27 '15 at 12:16
• @ClementC. - Can I ask how you got to $(k+1)k^2$? – RandomMath Apr 27 '15 at 12:21
• You're ultimately trying to show that $(k+1)! > (k+1)^2$, so think of working towards this, not from it. It should be your last line. – Théophile Apr 27 '15 at 12:24

$$(k+1)!> (k+1)^2$$ can be rewritten as
$$(k+1)\cdot (k!) > (k+1)\cdot(k+1)$$
For the induction step, here is the key part: \begin{align} (k+1)! &= (k+1)\cdot k!\tag{by definition}\\[0.5em] &> (k+1)\cdot k^2\tag{by the inductive hypothesis}\\[0.5em] &= k^3+k^2\tag{expand}\\[0.5em] &> k^2+2k+1\tag{since $k\geq 4$}\\[0.5em] &= (k+1)^2. \end{align} The main problem is realizing that you can deduce that $k^3+k^2>k^2+2k+1$ from the fact that $k\geq 4$.