Find the following limit: $\lim_{n\to \infty}\frac{n!e^n}{n^n}$. How can I find the  limit below: 
     $$\lim_{n\to \infty}\frac{n!e^n}{n^n}$$
I tried to use Stirling's approximation and got
$$
\lim_{n\to\infty}\frac{n!e^n}{n^n}=\lim_{n\to\infty}\frac{{\sqrt{2n\pi}(n/e)^n‌​}}{(n/e)^n}=\lim_{n\to\infty}\sqrt{2n\pi}=+\infty
$$
Is this right?

Let $$a_n =\frac{n! e^n}{n^n}$$ then $$\frac{a_{n+1}}{a_n} = e \left(1+\frac{1}{n}\right)^{n} \geq e \Rightarrow a_{n+1} \geq e\ a_n$$ Since $a_1=e$, $a_n \geq e^n \rightarrow \infty$, when $n\rightarrow \infty$, which means $\lim\ a_n = +\infty$
EDIT: Scratch that, I made a mistake, $$\frac{a_{n+1}}{a_n} \geq e \left( 1+\frac{1}{n} \right)^{-n}$$ Would someone be so kind as to downvote and/or delete this answer?
