I'm trying to figure out the asymptotic estimate of $\frac{(\sqrt{n}+n-1)!}{(\sqrt{n})!(n-1)!}$. I think it is $\omega(2^{\sqrt{n}})$ and am trying to prove it by showing $\lim\_{n \rightarrow \infty} \frac{\frac{(\sqrt{n}+n-1)!}{(\sqrt{n})!(n-1)!}}{2^{\sqrt{n}}} = \infty$ as wolfram says it should. But can't seem to get it. This is what I've tried so far:

$\frac{1}{2^{\sqrt{n}}}\frac{(\sqrt{n}+n-1)!}{(\sqrt{n})!(n-1)!} = \frac{1}{2^n}\frac{[\prod\_{i=1}^{n-1} (\sqrt{n}+i)](\sqrt{n})!}{(\sqrt{n})!(n-1)!} \\ = \frac{1}{2^n} \frac{\prod\_{i=1}^{n-1} (\sqrt{n}+i)}{(n-1)!}$

But I'm not sure where to go from here.