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.

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.
