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.
 A: 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) 
A: 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.
A: 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:
$$(1+n)^n=n^n+n\times n^{n-1}+\binom n2 n^{n-2} +\cdots>2n^n$$
A: We can induct on $n$. For $n=2$, it boils down to $4>3$. Now suppose it it true for $n=k$. Then we need to prove $k^k>(2k-1)!!\implies (k+1)^{k+1}>(2k+1)(2k-1)!!$
$$(k+1)^{k+1} =k^{k+1} +k\cdot k^k+ k(k-1)k^{k-1}/2 + \cdots > k^{k+1}+k\cdot k^k+(k^2-k)k^{k-1}/2=2.5k^{k+1} - 0.5k^k$$
by binomial theorem. Then $2.5k^{k+1}-0.5k^k = (k^k)(2.5k-0.5)>(2k-1)!!(2.5k-0.5)\geq(2k-1)!!(2k+1)= (2k+1)!!$
QED
