The problem:

Let $a\_{n} = 1 \cdot 3 \cdot 5\cdot . . . \cdot 2n-1 = \prod^{n}\_{k=1} \left(2k-1\right)$.

Then, use Stirling's formula to find $\lim\_{n\to\infty} \frac{a\_{n}}{\left(\frac{n}{e}\right)^{n} 4^{n} \sqrt{2}}$.

My work so far:

I know that $\prod^{n}\_{k=1} \left(2k-1\right)$ can be expressed as $\frac{\left(2n\right)!}{2^{n}n!}$ and I know that Stirling's formula is $n! \approx \left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$.

Now this is where I am stuck.

I have
$\lim\_{n\to\infty} \frac{\frac{\left(2n\right)!}{2^{n}n!}}{\left(\frac{n}{e}\right)^{n} 4^{n} \sqrt{2}}$.

Should I replace each $n!$ with $\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$, and then try and find the limit, or is there a more efficient way to solve this problem?