I was told to find a one-line proof for $n! \geq (\frac n e)^n$. I'm advised that Stirling's formula is not helpful. I've spent a little bit of time on it, but the solution is not coming to me. I feel like I'm missing an obvious trick. How do I do it?
With reference to Daniel Fischer's answer:
We know $e^n = \sum_{k = 0}^\infty \frac{n^k}{k!}$. That is, $$e^n = \frac{n^0}{0!} + \frac n 1 + \frac{n^2}{2} + \ldots + \frac{n^{n-1}}{(n-1)!}+ \frac{n^n}{n!}~.$$ Denote $$x = \frac{n^0}{0!} + \frac n 1 + \frac{n^2}{2} + \ldots + \frac{n^{n-1}}{(n-1)!}~.$$ Then $$e^n - x = \frac {n^n}{n!}~.$$ Clearly $x \geq 0$, therefore $$e^n \geq \frac{n^n}{n!} \Longleftrightarrow n! \geq \left(\frac n e \right)^n~.$$
