To prove by Mathematical Induction n!>n^2 for all n>3 My attempt:
Step 1  $n=4  \quad    LHS = 4! = 24   \quad  RHS=4^2 = 16$
Therefore $P(1)$ is true.
Step 2  Assuming $P(n)$ is true for $n=k, \quad k!>k^2, k>3$
Step 3 $n=k+1$
$LHS = (k+1)! = (k+1)k! > (k+1)k^2$ (follows from Step 2)
I am getting stuck at this step.
How do I show $(k+1)k^2 > (k+1)^2$ ?
 A: As $k+1>0$ it is sufficient to show $$k^2>k+1\iff k(k-1)>1$$ which is evidently true if $k\ge2$
Observe that $f(x)=x(x-1),$
$f(a)-f(b)=\cdot=(a-b)(a+b-1)$ will be $>0$ if $a>b$ and $a+b-1>0$
So, $f(x)$ will be increasing function for $a+b-1>0$
A: If you'd like to continue it your way, I believe you can finish with:
$(k+1)! = (k+1)k! > (k+1)k^2 = (k+1)((k+1)(k-1)+1) $
$= (k+1)^2(k-1) + (k+1)>(k+1)^2$ for all $k >3$
A: Core part of the induction proof:
$$
\begin{align*}
(k+1)!
&= (k+1)k! & \text{(factorial definition)}\\[1em]
&> (k+1)k^2 & \text{(by induction hypothesis)}\\[1em]
&> (k+1)(k+1) & \text{($k^2>k+1$ since $k\geq4$)}\\[1em]
&= (k+1)^2.
\end{align*}
$$
Note how the condition $k\geq4$ was used. 
A: $(k+1)k^2 ???? (k+1)^2$
$k^3 + k^2 ??? k^2 + 2k + 1$
$k^3 ??? 2k + 1$
Well,  .... that shouldn't be hard if $k > 3$
$k^3 = k^2*k > 9k = 2k + 7k > 2k + 21 > 2k + 1$.  (to be blunt)
.... 
If you want to be clever or creative:
$k^3 - 1 ??? 2k$
$(k-1)(k^2 + k + 1) ??? 2k$
$(k-1)(k^2 + k + 1) > 2(k^2 + k + 1) = 2k + 1 +(2k^2 + 1) > 2k + 1$.
But why be clever when you can be blunt.
