Prove $n! \geq n^2$ for $n \geq 4$

Question

I am working through a discrete math course, and have come upon a question that I don't understand how the solution was obtained.

The question is, prove $n! \geq n^2$

Hypothesis: $p(n): n! \geq n^2, n\geq 4$

Basic step: $p(4): (4)! \geq (4)^2$. $24 \geq 16$, so $p(4)$ is true.

Prove $p(n+1): (n+1)! \geq (n+1)^2$

(left side)

$n!(n+1)$ (because $n!(n+1)$ is the same as $(n+1)!$)

$n^2(n+1)$ (by inductive hypothesis)

$n^3+n^2 $

... and here is where I get stuck. The solutions key continues as follows: $n^3+n^2 \geq n^3 + 3n^2$ (I don't understand where the $3n^2$ comes from) $= n^3 + 2n^2 + n \geq n^2 + 2n + 1 = (n + 1)^2$

I understand how $n^2 + 2n + 1 = (n + 1)^2$, I even get how $n^3 + 2n^2 + n \geq n^2 + 2n + 1$ by factoring out $n$... but I don't understand where $n$ goes. Shouldn't it read $n^3 + 2n^2 + n \geq n(n^2 + 2n + 1)$ ?

Thanks for any help offered,

The solution key (as you've presented it) is incorrect: $n^3+n^2$ is less than $n^3+3n^2$, not greater than. — Barry Cipra, Feb 20 '14 at 16:06
Not proposing to close, just linking to the node of this network of duplicates. — Lee David Chung Lin, Feb 06 '20 at 18:51

Barry Cipra · Accepted Answer · 2014-02-20T16:28:24.963

There's something not right in the solution key (as presented). Once you've used the induction hypothesis to establish $(n+1)!\ge (n+1)n^2$, all you have to show is that $(n+1)n^2\ge(n+1)^2$. This amounts to showing $n^2\ge n+1$, which is true as soon as $n\ge2$, since $n^2\ge2n=n+n\gt n+1$.

It's possible the solution key meant to say

$$n^3+n^2\ge n^2+3n = n^2+2n+n\ge n^2+2n+1=(n+1)^2$$

in which case the first step combines $n^3\ge n^2$, which is true for all (non-negative) $n$, and $n^2\ge3n$, which is true for $n\ge3$.

Prove $n! \geq n^2$ for $n \geq 4$

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