Prove the inequality for n>2. Show that $(n!)^2>n^n$ for
$n \in \Bbb Z_+$.
I tried induction but I was getting stuck in the last step.
Please help. 
Hints only, please.
 A: We have 
$$(n!)^2 = \prod_{a=1}^n a(n+1-a) > \prod_{a=1}^n n = n^n,$$
because $a(n+1-a) \geq n$ for all $1 \leq a \leq n$. The latter inequality is equivalent to $(n-a)(a-1) \geq 0$ which is obviously true and also strict when $a=2$, $n>2$.
A: HINT:
Write $1$ to $n$ in a line and below that write $n$ to $1$. Now multiply the two no.s in each columns. 
A: Okay since you've said you tried induction, I'm assuming you reached the step
$(k!)^2 * (k+1)^2 > k^k*(k+1)^2 $
Now if you can show that 
$k^k*(k+1)^2>(k+1)^{k+1}$
then you're done.
Why not try keeping the (k+1) terms in the right hand side?
Now we arrive at
$k^k>(k+1)^{k-1}$
Both the terms we have are in the form of products of integers, it might be easier if only one side is a product.(Why?)
So we can rewrite it again as
$1>\frac{1}{k}*(1+\frac{1}{k})^{k-1}$.
I hope now it is easy to see how to proceed.
A: For a help on how to prove the problem by Induction suppose that the inequality is true for all $n>1$ , i.e, we have (for that $n$) $(n!)^2$ $>$ $n^n$ $\implies$ $\dfrac{(n!)^2}{n^n}$ $>$ $1$. Let $u_n$ $=$ $\dfrac{(n!)^2}{n^n}$. Then try to prove that $u_n$$_+$$_1$ $>$ $u_n$ for all $n>1$. 
