Limit similar to $\lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \text{e}^{-1}$

We have, by a simple comparison, $ \left( 1-\frac{n}{n^2-n} \right)^n \le \prod_{i=0}^{n-1} \left( 1-\frac{n}{n^2-i} \right) \le \left( 1-\frac{1}{n} \right)^n $. The limit of the left and right hand products as $ n \rightarrow \infty $ is $ \frac{1}{e} $, so by the Squeeze theorem, the limit of the middle product is also $ \frac{1}{e} $ as you wanted.