Solving $\lim\limits_{n\to\infty}\frac{n^n}{e^nn!}$ I was solving a convergence of a series and this limit popped up:
$$\lim\limits_{n\to\infty}\frac{n^n}{e^nn!}$$
I needed this limit to be $0$ and it is in fact (according to WolframAlpha), but I just don't see how to get the result.

Almost as good: write the expression as 
$$
L = e^{n \log n - n - \log n!} = e^{n \log n -n -\sum_{k=1}^{n} \log k} 
$$
and use the bounds on the sum:
$$
\int_{1}^{n} \log x dx < \sum_{k=1}^{n} \log k < \int_{1}^{n+1} \log x dx
$$
to get the same result without Stirling. 
