Is there any tight lower bound on following expression for $n\geq k$
$$n(n-1)(n-2) \dots (n-k+1)$$
one lower bound I could think is $(n-k+1)^k$. But I need to find $f$ s.t
$$f(k) n^{k} \leq n(n-1)(n-2) \dots (n-k+1)$$
my attempt using stirling's approximation
$$n(n-1)(n-2) \dots (n-k+1) = k! \binom{n}{k} \geq \sqrt{2\pi}k^{k} \sqrt{k} e^{-k} . (n/k)^{k} = \Big[\sqrt{2\pi k} e^{-k}\Big] n^k $$. I need to know if I can remove $e^{-k}$ in above and get something more like $k^{-c}$, atleast when $k$ is small. Any pointers and suggestions would be helpful.
$$n(n-1)(n-2)\cdots(n-k)=\frac{n!}{(n-(k+1))!}=\binom{n}{k+1}(k+1)!$$
Now we know that $$\left(\frac{n}{k+1}\right)^{k+1} \leq {n \choose k+1} \leq \left( \frac{en}{k+1}\right)^{k+1}$$ Therefore we get that $$\left(\frac{n}{k+1}\right)^{k+1}(k+1)!\leq n(n-1)(n-2)\cdots(n-k)\leq\left( \frac{en}{k+1}\right)^{k+1}(k+1)!$$ where $e=\sum_{m=0}^{\infty}\frac{1}{m!}$