Hint in Proving that $n^2\le n!$ The problem is to find the values of n such $n^2\le n!$.
I have found this set to be {${n\in Z^+| n\le1 \lor n\ge 4}$}
I've used a proof by cases to prove the first part ($\forall n\le 1$) which was quite simple. I know I need to use mathematical induction. I've verified the basis step but I don't know how to go about proving the inductive hypothesis. Simply writing $(k+1)^2\le(k+1)!$ is simply affirming the conclusion and I can't see how to manipulate this. Could someone give me a hint on what to do here? I'd prefer to not get a full answer, just a hint. 
Thanks in advance.
 A: There is no real need to use induction. Note that if $n\ge 3$ then $n!\ge (n)(n-1)(n-2)$.
To prove that $n!\ge n^2$, it is enough to prove that $(n-1)(n-2)\ge n$, or equivalently that $n^2-4n+2\ge 0$. Since $n^2-4n+2=n(n-4)+2$, this is obviously true if $n\ge 4$. 
A: The induction hypothesis is assumed.
Induction hypothesis: Assume $k^2 \leq k!$.
Now you use the inductive hypothesis to prove the inductive step:
$$(k+1)^2 = k^2 + 2k + 1 \overset{?}{\leq} (k+1)! = (k+1)k!= kk! + k!$$
Can you see how to confirm that the inequality holds?
A: As a way of working a problem like this in general, you can write the expression $(k + 1)^2$ at the left end of a line of your paper and the expression $(k + 1)!$ on the right end of the same line. But as you say, you cannot yet write $\le$ between them.  (You might instead write a questoin mark.)  Now underneath each of these expressions, start writing other expressions that are equal to them (other answers have already given expressions you will find useful, but eventually you'll be able to intuit things to try on your own) until you find that you can show that the left-side expression is no greater than the right-side expression.  This might mean you eventually have something of the form $k^2 + A$ on the left and $k! + B$ on the right, and you know for other reasons that $A \le B$.  Since you already know that $k^2 \le k!$, this is enough to prove your statement.
A: Hint: $$(k+1)!=(k+1).k!$$ ${}{}{}{}{}{}$
A: $$n!=(n^2-n)(n-2)!=\prod\limits_{i=0,\\\,i\, \mathrm{even}}^{n-2}\left(n^2-(2i+1)n+i(i+1)\right)$$
A: The basic step is trivial: check whether the inequality holds for $n \ge 4$:
$4^2 = 16 \le 4! = 24$
Once done this, let's go on to the inductive step, which consists of considering the inequality true for an integer $n$, and verifying whether it is the case for an integer $n+1$:


This holds true as stated by the hypothesis: $n^2 \le n!$.
Now, the inequality we ought to prove: $(n+1)^2=$
$=n^2 + 2n + 1$
$\le \underbrace{n!}_{hypothesis} + 2n + 1$
I can sum $n!$ $n$ times on the right hand, and still hold the inequality true,
$\le n! + \underbrace{n!+...+n!}_{n \space times} + 2n + 1$
$\le n!(n+1) + 2n + 1$

$2n+1$ can be eliminated, since $n!(n+1)$ is greater.
$\le (n+1)!$
QED

A: For $n\ge 4\,,$ $n!\ge n(n-1)(2)$ and $(n-1)(2)\ge n\,.$ $k=1,2,3$ you can check by hand.
