Evaluate $\lim_{n\to\infty} \frac{n!e^n}{n^n}$ with L'Hopital's rule (or without) 
The problem is to find the limit
$$\lim_{n\to\infty} \frac{n!e^n}{n^n}.$$

My first idea was reorder terms:
$$\lim_{n\to\infty}\frac{n!e^n}{n^n}=\lim_{n\to\infty} n!\left(\frac{e}{n}\right)^n$$
with the indeterminate form $\infty \cdot 0$. Reordering terms:
$$\lim_{n\to\infty} \frac {\left(\frac{e}{n}\right)^n}{1/n!}$$
with the form $0/0$. Can I apply L'Hopital's rule to evaluate the limit? Using WolframAlpha, the answer is that the limit don't exist ($\infty$):
https://www.wolframalpha.com/input?i=limit+x-%3E+inf+%28x%21e%5Ex%2Fx%5Ex%29,
but i can't find the way to find this answer!!
 A: First of all, L'Hopital's rule is for some differentiable function. Sure, you can do it if you replace the factorial with the Gamma function, but this is a bit annoying.
Anyway, another way to do this: denote by $ a_n := \dfrac{n!e^n}{n^n} $. Then we have the ratio
$$ \frac{a_{n+1}}{a_n} = \frac{e}{\left(1 + \frac{1}{n}\right)^n} .$$
Take logs and you get
$$ \log{a_{n+1}} - \log{a_n} = 1 - n\log\left(1 + \frac{1}{n}\right) = \frac{1}{2n} + O\!\left(\frac{1}{n^2}\right)$$
using the Maclaurin expansion of $\log(1 + x)$.
We then have that $\log{a_n} \to \infty$ by telescoping and the divergence of the harmonic series, and so $a_n \to \infty$ as well.
A: HINT
Note that if you take logs,
$$
\begin{split}
L &= \lim_{n\to\infty} \left(\frac{n!e^n}{n^n}\right) \\
  &= \exp\left(\ln\left(\lim_{n\to\infty} \frac{n!e^n}{n^n}\right)\right) \\
  &= \exp\left(\lim_{n\to\infty} \ln\left(\frac{n!e^n}{n^n}\right)\right) \\
\end{split}
$$
Can you finish?
A: The limit can simply be evaluated using Stirling's fomula. About this fomula, you can see Section 8.22 of Baby Rudin.
Since $\frac{\Gamma\left(x+1\right)}{\left(x/e\right)^x\sqrt{2\pi x}}\to 1$ as $x\to \infty$, note that $\Gamma\left(n+1\right)=n!$ when $n\in\mathbb{N}$, obviously the limit in your problem is $\infty$.
A: I will make use of $$\left (1+{1\over n}\right )^n\nearrow e, \qquad (1+x)^n\ge 1+nx,\ x\ge 0$$
Let $a_n=\displaystyle{n!e^n\over  n!}.$ Then
$${a_{n+1}\over a_n}={e\over \left (1+{1\over n}\right )^n}\ge {\left (1+{1\over 2n}\right )^{2n}\over \left (1+{1\over n}\right )^n}={\left (1+{1\over n}+{1\over 4n^2}\right )^n\over \left (1+{1\over n}\right )^n}\\ =
\left ( 1+{1\over 4n(n+1)}\right )^n\ge 1+{1\over 4(n+1)}$$
Hence
$$a_n\ge a_1\prod_{k=2}^n \left (1+{1\over 4k}\right )\ge e\left [1+\sum_{k=2}^n{1\over 4k}\right ]$$
Therefore $\displaystyle \lim_n a_n=\infty.$
