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.

For the first few cases this checks out, then assume it is true for $n$
$$n^n>(2n-1)!!$$
Then
$$(n+1)^{n+1}=\color{blue}{(n+1)^n (n+1)>2n^{n}(n+1)}=(2n-1)!!\times (2n+2)>(2n-1)!!\times (2n+1)>(2n+1)!!$$
Using the binomial theorem for the blue part:
$$(1+n)^n=n^n+n\times n^{n-1}+\binom n2 n^{n-2} +\cdots>2n^n$$
