Limit $lim (frac{n!}{n^n})^{frac{1}{n}}$

I need to calculate $$\lim_{n\rightarrow \infty} \left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$ My try:

When $n!$ is large we have $n!\approx(\frac{n}{e})^n\sqrt {2\pi n}$ (Stirling approximation) $$\lim_{n\rightarrow \infty} \left(\frac{n!}{n^n}\right)^{\frac{1}{n}}=\lim_{n\rightarrow \infty}\frac{\left((\frac{n}{e})^n\sqrt {2\pi n}\right)^{\frac{1}{n}}}{n}$$ Simplifying we get, $$\frac{1}{e}\lim_{n\rightarrow \infty} \left(\sqrt{2\pi n}\right)^{\frac{1}{n}}$$

I am stuck here. I don't know how to proceed further.

Answer

Another way :

Let $$A= \lim_{n \to \infty}\left(\frac{n!}{n^n}\right)^{\frac1n}$$

$$\implies \ln A= \lim_{n \to \infty}\frac1n\sum_{1\le r\le n}\ln \frac rn$$

Now use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$