Using Stirling's Formula

Question

I am trying to use Stirling's formula to show that

$${{2n} \choose {n}} \cdot .5^n \cdot .5^{2n-n} \approx \frac{1}{\sqrt{\pi n}}$$

Stirling's Formula is $n!=n^ne^{-n}\sqrt{2\pi n}$

I have attempted the below and am having a hard time showing this.

$${{2n} \choose {n}} \cdot .5^n \cdot .5^{2n-n}=$$

$$\frac{2n!}{n!(2n-n)!} \cdot .5^n \cdot .5^{n}=$$

$$\frac{2n!}{(n!)^2}\cdot .5^n \cdot.5^n=$$

$$\frac{2}{n!}\cdot .5^n \cdot.5^n = $$

Here I substitute in for the formula

$$\frac{2}{n^ne^{-n}\sqrt{2\pi n}}\cdot .5^n \cdot.5^n =$$

$$\sqrt{\frac{2}{\pi}}e^nn^{-n-\frac{1}{2}}\cdot .5^n \cdot.5^n$$

I do not think this equals what I am trying to show and I do not know where to go from here.

Answer 1

It is $\dfrac{(2n)!}{n!n!}$, not $\dfrac{2(n)!}{n!n!}$.