Prove that $n^n>1\times3\times5\times7\times \dots\times(2n-1)$ The main question is :
Prove that $n^n>1\times3\times5\times7\times\dots\times(2n-1)$
My approach :
We can write the R.H.S as, 
$$\frac{(2n-1)!}{2\times4\times6\times\dots\times2(n-1)}$$
We can write $(2n-1)!$ as $(2n-3)!(2n-1)2(n-1)$
Thus, in this manner we can cancel out even terms till $n$ appears. I'm having trouble doing this. What I intend to do is say that any positive number's factorial is less than the number raised to itself. I welcome any alternate method. Please help.
P.S. I am still in high school and do not understand binomial theorem and induction.

HINT:
Using AM, GM inequality for $r,2n-r>0$ 
$$\dfrac{r+(2n-r)}2\ge\sqrt{r(2n-r)}$$
Set $r=1,3,5\cdots, 2n-3,2n-1$  and multiply.
Observe that the equality can not be held  as $r=2n-r$ will not occur unless $r=n$(which is again possible if  $n$ is odd) 
