Math induction ($n^2 \leq n!$) help please I'm having trouble with a math induction problem.  I've been doing other proofs (summations of the integers etc) but I just can't seem to get my head around this.

Q.  Prove using induction that $n^2 \leq n!$

So, assume that $P(k)$ is true: $k^2 \leq k!$
Prove that $P(k+1)$ is true: $(k+1)^2 \leq (k+1)!$
I know that $(k+1)! = (k+1)k!$ so:  $(k+1)^2 \leq (k+1)k!$  but where can I go from here?
Any help would be much appreciated.
 A: The statement you want to prove is for all $n\in\mathbb{N}$ it holds that $n^2\leq n!$ (you called this $P(n)$. So lets first prove $P(4)$ i.e. $4^2\leq 4!$ but since $16\leq 24$ this is clear. So lets assume $P(n)$ and prove $P(n+1)$.
First note that for $n\geq 2$ it holds that 
$$ 0\leq (n-1)^2+(n-2)=n^2-2n+1+n-2=n^2-n-1 $$
which is equivalent to $n+1\leq n^2$ which gives
$$ (n+1)^2=(n+1)(n+1)\leq (n+1)n^2 $$
by induction hypothesis (i.e. $P(n)$) the term $n^2$ in the last expression is smaller or equal $n!$ so we can continue:
$$ (n+1)n^2\leq (n+1)n! = (n+1)! $$
which is the statement we wanted to prove.
My answer is very extensive and explicit. But maybe, you now get a better understanding of what you have to do in general, when you want to prove something by induction.
A: Suppose $k^{2}\leq k!$. Then $(k+1)!=(k+1)k!\geq (k+1)k^{2}\geq (k+1)^{2}$, whenever $k\geq 2$.
A: One way to prove this is to strengthen your induction hypothesis slightly: assume $k^2+k\leq k!$ instead of $k^2\leq k!$. You have a base case for this starting with $k=4$, where $k^2+k=20$ and $k!=24$. 
Then we want to show $(k+1)^2+k+1<(k+1)!$. Divide both sides by $k+1$ to see this holds just if $k+2<k!$. But we supposed $k^2+k<k!$ and $k\geq 4$, so this holds. Then we've shown for all $n$ that $n^2+n\leq n!$, which implies the conclusion.
