Solution: Let $x_n = \frac{n!}{n^n}$. Note that for $n\ge 2$ \begin{align*} \frac{n!}{n^n} & = \frac{1\cdot 2\cdot 3\cdot 4\cdot \cdots \cdot n}{n\cdot n\cdot n\cdot n\cdot \cdots \cdot n} \\[1ex] & = \left( \frac{1}{n} \right) \left( \frac{2}{n} \right) \left( \frac{3}{n} \right) \cdots \left( \frac{n}{n} \right) \\[1ex] & \leq \left( \frac{1}{n} \right) \left( \frac{n}{n} \right) \left( \frac{n}{n} \right) \cdots \left( \frac{n}{n} \right) \\[1ex] & = \frac{1}{n}. \end{align*} Therefore, we have \[ 0 \leq x_n \leq \frac{1}{n}. \] By the Sandwich's theorem, we conclude that \[ \lim_{n \to \infty} \frac{n!}{n^n} = 0. \]