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Show that $$n^{(l)}=n(n-1)\dots(n-l+1)\ge {n^l\over e}$$ where $2\le l \le \sqrt{n}$

Here is how far I've got with this:

$$n^{(l)}=\prod_{i=0}^{l-1}(n-i)=n^l\prod_{i=1}^{l-1}(1-{i\over n})\\ \ge n^l (1-{l-1\over n})^{l-1}\ge n^l (1-{l-1\over n})^{n} $$

Unfortunately this is not in the right form to apply a bound on the exponential function. Also I am worried that I haven't used the constraint on $l$ either.

Could anyone help me with this?

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up vote 1 down vote accepted

By convexity, on $0\le x\le \frac12$, $1-x\ge e^{-2x}$. Then after your first line, observe that in the product, $i\le \sqrt{n}-1\le \frac n2$ and so $$ \prod_{1\le i\le l-1} (1-\frac in)\ge \prod_{1\le i\le l-1} e^{-2i/n} = \exp -\frac 2n \frac{l(l-1)}{2}\ge \exp -1. $$

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