Prove that $\lim_{n \to \infty} \frac{n!}{n^n} = 0$ Prove that


  $\lim_{n \to \infty} \frac{n!}{n^n} = 0$ 


I've already considered using l'Hoptials rules but I cannot take the derivative of a factorial (as it is a discrete function). 

Thanks
The easiest way is to get an explicit upper bound for the fraction:

$0 &lt; \dfrac{n!}{n^n} = \dfrac{1\cdot 2\cdot ...\cdot n}{n\cdot n\cdot n\cdot...\cdot n} &lt; \dfrac{1}{n}$, and $\dfrac{1}{n} \to 0$. So the answer follows by squeeze theorem.
Another approach: Stirling's approximation tells you $n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n$. That tells you the thing you're taking the limit of is approximately $\frac{2 \pi n}{ e^n}$. This goes to 0 as $n \to \infty$. 

Alternatively, you can use the bound from Stirling's approximation is $n! \leq e \sqrt{n} n^n e^{-n}$. Then, $0 \leq \frac{n!}{n^n} \leq e \sqrt{n} e^{-n} \to 0$.