Prove by induction $n! > n^2$ for $n \geq 4$

Question

This theorem is supposed to be true for $n \geq 4$

I’ve tried $n!(n+1) > n^2(n+1) $

Not sure where to go from here or if I’m on the right track

Answer 1

Hint: Multiply your inequality $$n!>n^2$$ by $$n+1>0$$ then you will get

$$(n+1)!>n^2(n+1)$$ and show that $$n^2(n+1)>(n+1)^2$$

Answer 2

HINT

Recall that we need to proceed by

• base case: $n=0 \implies 0!=1>0^2$
• induction step: assuming true $n!>n^2$ we need to prove $(n+1)!>(n+1)^2$ then

$$(n+1)!=(n+1)n!\stackrel{Ind. Hyp.}>(n+1)n^2\stackrel{?}>(n+1)^2$$

therefore it reduces to prove that

$$(n+1)n^2>(n+1)^2$$

Note that at the end if necessary we need to revise the base case.