Formal proof of the sequence involving double factorials $x_n = \frac{(2n)!!}{(2n-1)!!}$ is not bounded. I'm trying to prove the following:

Let $n\in \mathbb N$ and 
  $$ 
x_n = \frac{(2n)!!}{(2n-1)!!}
$$ 
  Where:
  $$
(2n)!! = 2\cdot 4 \cdot 6 \cdot \dots \cdot 2n \\
(2n - 1)!! = 1\cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)
$$
  Prove $\{x_n\}$ is not bounded.

Intuitively it feels like it is unbounded. Let's consider the following fraction:
$$
x_n = \frac{2\cdot 4 \cdot 6 \cdot \dots \cdot (2n)}{1\cdot 3 \cdot 5 \cdot \dots \cdot (2n-1) }
$$
Now if we consequently take a number from nominator and denominator we get:
$$
\frac{2}{1} > 1 \\
\frac{4}{3} > 1 \\
\frac{6}{5} > 1 \\
\dots \\
\frac{2n}{2n-1} > 1
$$
So the fraction is a product of rational numbers each of which is greater than $1$ and the product of rational numbers greater than $1$ is increasing. 
I've tried to formalize that by expanding $(2n)!!$ and $(2n-1)!!$:
$$
(2n)!! = (2n)(2n-2)(2n-4)\cdots(4)(2) = 2^kn!\\
(2n-1)!! = (2n-1)(2n-3)\cdots(5)(3)(1) = \\
= \frac{(2n-1)(2n-2)(2n-3)\cdots(5)(4)(3)(2)(1)}{(2n-2)(2n-4)\dots(4)(2)} = \\
\frac{(2n-1)!}{2^{n-1}(n-1)!}
$$
So using the above:
$$
x_n = \frac{2^nn!2^{n-1}(n-1)!}{(2n-1)!} = \frac{2^{2n-1}n!(n-1!)}{(2n-1)!}
$$
Here is where I got stuck. How do i proceed with the proof using some constant $M$ and some number $N > n$ such that $x_N > M$? Should i introduce some inequality?
I've seen a similar question, but the sequence there is proven to be divergent which i guess is more a calculus concept (yet very similar to (un)boundedness), and i'm in search of a precalculus solution.

Without integrals or derivatives:
$$ x_n=\frac{(2n)!!}{(2n-1)!!}= \prod_{k=0}^{n-1}\left(1+\frac{1}{2k+1}\right)=\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^{-1}$$
implies
$$ x_n^2 = \prod_{k=1}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right)^{-1}=\frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)^{-1}\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right) $$
or
$$ x_n^2 = \frac{n}{4}\prod_{k=1}^{n-1}\left(1-\frac{1}{(2k+1)^2}\right). $$
On the other hand the infinite product $\prod_{k\geq 1}\left(1-\frac{1}{(2k+1)^2}\right)$ is convergent to a positive constant since $\sum_{k\geq 1}\frac{1}{(2k+1)^2}$ is convergent to a positive constant, hence $x_n\geq \sqrt{K n}$ for some mysterious constant $K>0$, which actually equals $\pi$.
