Prove that for any $n \ge 2$,$1\times3\times5\times \dots \times(2n-1)Prove that for any $n \ge 2$,$1\times3\times5\times \dots \times(2n-1)<n^n$ without induction
I asked for a non induction prove but I am stuck in induction prove too.
In induction we should prove $1\times3\times \dots\times(2n+1)<(n+1)^{n+1}$ therefore we should prove $\cfrac{(n+1)^{n+1}}{n^n}>2n+1$ but I cannot prove it please post both induction and non induction proves.
 A: Non-Inductive Proofs:
$(1)$ Observe that 
$$1\cdot (2n-1)<n^2$$
$$3 \cdot (2n-3)<n^2$$
$$\ldots$$
$$\ldots$$
In general,
$$r\cdot(2n-r) <n^2$$
Multiplying all these inequalities we get the desired result, when $n$ is even.
When $n$ is odd, we need to multiply both sides by $n$ to get the desired result.
$(2)$ Just as Zain Patel commented:
$$1 \cdot 3 \cdots (2n-1)=(2n-1)!!=\frac{(2n)!}{2^n \cdot n!}$$
Now, we are left to prove 
$$\frac{(2n)!}{2^n \cdot n!} <n^n \Longleftrightarrow (2n)! < (2n)^n \cdot n!$$
Observe that $$(2n)!=n! \cdot \prod_{i=1}^{n}{n+i}$$
So, we are left to prove $$\prod_{i=1}^{n}{(n+i)}<(2n)^n$$ which is obvious.

Inductive Proof:
Hand-waving that the Base Case is true and continuing from where you left.
$$1 \cdot 3 \cdot 5 \cdots (2n-1) \cdot (2n+1) <n^n \cdot(2n+1) \tag{Inductive Hypothesis}$$
So, we are left to prove 
$$(n+1)^{n+1} > n^n \cdot (2n+1) \Longleftrightarrow \left(1+\frac{1}{n}\right)^n > \frac{2n+1}{n+1}=2-\frac{1}{n+1}\tag{1}$$
Using Bernouli's inequality, we have 
$$\left(1+\frac{1}{n}\right)^n > 1+\left(n\cdot \frac{1}{n}\right)=2\tag{2}$$
From $(1)$ and $(2)$, we get the desired result.
A: Just as Zain Patel commented $$1\cdot 2 \cdots (2n-1) = (2n-1)!! = \frac{(2n)!}{2^n n!}$$ Now, using Stirling approximation
$$k! \sim \sqrt{2\pi k}\left(\frac k e \right)^k$$ $$\frac{(2n)!}{2^n n!}\sim2^{n+\frac{1}{2}} e^{-n} n^n=\sqrt{2}\left(\frac 2 e \right)^n n^n$$ and, since $e>2$, for any $n>2$, the result.
