I am trying to prove the following inequality,
$$n^{(l)} = n(n-1)\cdots(n-l+1) \geq \frac{n^l}{e}\quad\text{ for }\quad 2 \leq l \leq \sqrt{n}\;.$$ So my approach is to observe that $n^{(l)} = \dfrac{n!}{(n-l)!}$ and apply Stirling's Approximation, to get
$$n^{(l)} \geq \frac{\sqrt{2\pi n}n^ne^{-n}}{e\sqrt{n-l}(n-l)^le^{l-n}}\;.$$ So in order to get to the desired inequality it suffices to show that for $2 \leq l \leq \sqrt{n}$, $$\sqrt{2\pi}n^{n-l+1/2} \geq e^l(n-l)^{n-l+1/2}\;,$$ but I'm not sure what to do or whether my approach is optimal. It seems like this is the right approach as it would allow me to find the condition on $l$.