Calculating $\lim\limits_{n\to\infty}\frac{n!\cdot e^n}{n^n}$ I tried using the same trick as $\lim\limits_{n\to\infty}\frac{n!}{n^n}$, where you compare the terms one to one.
$(\frac{1}{n})(\frac{2}{n})(\frac{3}{n})...(\frac{n}{n})\cdot e^n$ = $(\frac{e}{n})(\frac{2e}{n})(\frac{3e}{n})\cdots(\frac{ne}{n})$
I can't really figure out what to do after this.  Some of the terms are less than one and some are greater than one.  It would be great if someone could lead me in the right direction of what to do here.
 A: Apply Stirling's approximation so that as $n\to\infty$ 
$$n! \sim \left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$$
therefore
$$\frac{n!\cdot e^n}{n^n}\sim \sqrt{2\pi n}$$
from which $\sqrt{2\pi n}$ increases monotonically as $n\to\infty$
$$\lim\limits_{n\to\infty}\sqrt{2\pi n}=\infty $$
thus
$$\lim\limits_{n\to\infty}\frac{n!\cdot e^n}{n^n}=\infty$$
A: Your are going in right direction
See as n goes to infinity n>>> e so the first term becomes very less.now for the last term n would cancel out itself so last term is e
So limit equals to 0 because maximum value that is multiplied is e and every other value is very close to zero
A: Use Sterling formula and you have $$\frac {n! e^n}{n^n}\approx \sqrt {2\pi n}(n/e)^n (e/n)^n = \sqrt{2\pi n} \to \infty$$
A: I think another way to go about this would be to use the trick you were initially planning to and combine it with the taylor expansion of $e^n$. Then you will have:
$$ \lim _{n \rightarrow \infty} \frac{n ! \cdot e^{n}}{n^{n}} = \lim _{n \rightarrow \infty} \frac{n !}{n^{n}} \cdot (1+n+\frac{n^{2}}{2 !}+\frac{n^{3}}{3 !}+\frac{n^{4}}{4 !}+\frac{n^{5}}{5 !}+\dots)$$
and you can open the brackets and apply the trick you mentioned.
