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.

Since $(2n-1)!!=\frac{(2n)!}{2^n n!}$, your inequality is equivalent to
$$(2n)! < n! (2n)^n \tag{1}$$
or to:
$$ \underbrace{(2n)\cdot(2n-1)\cdot\ldots\cdot(n+1)}_{n \text{ terms}}<\underbrace{(2n)\cdot(2n)\cdot\ldots\cdot(2n)}_{n \text{ terms}}\tag{2} $$
that is trivial.
