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?

Using Stirling's Approximaton, as $n\to\infty$:
$$n! \sim \left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$$
Therefore:
$$\frac{n!e^{n}}{n^{n}}\sim \sqrt{2\pi n}$$
Which clearly, as $\sqrt{kn}$ is monotonically increasing $\forall n>0$, blows up to infinity as $n\to\infty$, we therefore have:
$$\lim_{n\to\infty}\frac{n! e^{n}}{n^n}=\infty$$
