Range of $N=\frac21\times\frac43\times\dots\times\frac{400}{399}$? Question:

If $N=\frac21\times\frac43\times\dots\times\frac{400}{399}$ then it lies between:
a) $50$ and $60$
b) $20$ and $30$
c) $30$ and $40$
d) $40$ and $50$

What I tried:

$M1$: Tried finding of someway to reduce the expression. Taking $2$ common from all the terms in numerator didn't help. Then writing terms as $(1+\frac1{2k-1})$ also didn't help.
$M2$: Multiplied first few terms to check if I get somewhere close to the answer as the latter terms are very small. But nope, first few terms didn't really give me a nice idea.

Please help.
 A: It is a well known fact that
$$\frac 1{\sqrt{4n+2}}\le \frac{1.3.5\dots (2n-1)}{2.4.6\dots (2n)}\le \frac 1{\sqrt{3n+1}}$$
It can be proven by induction or by noticing the fact that
$$\frac ab\le \frac{a+k}{b+k}$$
and then taking a clever product.
A: $$N = \frac{\left(2\times4\times...\times400  \right)^2}{400!} =\frac{\left(2^{200}200! \right)^2}{400!}$$
Applying the Stirling's formula:
$$n! \approx \sqrt{2\pi n}\left(\frac{n}{e}  \right)^n$$
Then
$$\color{red}{N} \approx\frac{\left(2^{200}\sqrt{2\pi \times200}\left(\frac{200}{e}  \right)^{200} \right)^2}{\sqrt{2\pi \times400}\left(\frac{400}{e}  \right)^{400}}=\sqrt{2\pi}\times 10 \color{red}{\approx25.07}$$
Then, we should choose $\color{red}{\mathbf{b}}$: $N$ lies between $20$ and $30$.
Q.E.D
A: Here's what I got from a friend:
First we find the lower limit of $N$.
We know that $\frac21>\frac32,\frac43>\frac54,\dots\frac{398}{397}>\frac{399}{398} \ \text{and }\frac{400}{399}>1$
Let $X=\left(\frac32\right)\left(\frac54\right)\dots\left(\frac{399}{398}\right)(1)$
$NX=400$. Also, $N>X$
So, $\bf N>20$
Now we'll find the upper limit of $N$.
$N^2=\left(\frac21\right)^2\left(\frac43\right)^2\left(\frac65\right)^2\dots\left(\frac{400}{399}\right)^2$
Let $Y=\left(\frac21\right)^2\left(\frac32\times\frac43\right)\left(\frac54\times\frac65\right)\dots\left(\frac{399}{398}\times\frac{400}{399}\right)$
$N^2<Y$. Also, $Y=(2)(400)\Rightarrow N^2<(2)(400)$
So, $\bf N<20\sqrt2\sim28.2$
Thus, $\bf N$ lies between $\bf 20$ and $\bf 30$.
