I'm trying to estimate $\binom{n}{\left \lfloor{\alpha n}\right \rfloor }$ asymptotically using Stirling's formula. However, I'm a little lost with what to do about the floor function here.
In the case without the floor function, there is a greater ease to combine the $\alpha$ and $n$ terms, such as $$\binom{n}{\alpha n}=\frac{n!}{\alpha n!(n-\alpha n)!} \sim \frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{\sqrt{2\pi \alpha n}\left(\frac{\alpha n}{e}\right)^{\alpha n} \sqrt{2\pi (n-\alpha n)}\left(\frac{(n-\alpha n)}{e}\right)^{(n-\alpha n)}} =\frac{1}{\sqrt{2 \pi\alpha (n- \alpha n)} \alpha^{\alpha n}\left(\left(\frac{n}{e}\right)^{n}\right)^{(\alpha - 1)} \left(\frac{(n-\alpha n)}{e}\right)^{(n-\alpha n)}}$$ I'm not sure if there's a further simplification of this (if there is, please left me know!) but I'm also not sure how this would work with the floor function.
I tried splitting into cases, so if $\frac{1}{\alpha}<n$, then $\left \lfloor{\alpha n}\right \rfloor=0 \implies \binom{n}{\left \lfloor{\alpha n}\right \rfloor}=n!$ and when $\frac{1}{\alpha}=n$, then $\left \lfloor{\alpha n}\right \rfloor=1 \implies \binom{n}{\left \lfloor{\alpha n}\right \rfloor}=n!$. However, the case when $\frac{1}{\alpha}>n$ leaves $\left \lfloor{\alpha n}\right \rfloor$ the same.
The asymptotic expansion you have written without the floor function implicitly uses the floor function or you couldn't have approximated $(\alpha n)!$.
What you have done is a good start. The next thing to do is replace all occurrences of $n-\alpha n$ by $n(1-\alpha)$ so you can group more terms with $n$ together.
I'll let you do that. If you still have problems, comment and I'll proceed further.