Limit similar to $\lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \text{e}^{-1}$ I want to show that 
$$
\lim_{n \to \infty} \left(1-\frac{n}{n^2} \right) \left(1-\frac{n}{n^2-1} \right)  \cdot \ldots \cdot \left(1-\frac{n}{n^2-n+1} \right)
= \lim_{n \to \infty} \prod_{k=0}^{n-1}
\left(1-\frac{n}{n^2 - k} \right)
= \mathrm{e}^{-1}
$$
I know that $\lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \text{e}^{-1}$ and so my attempt was to write 
$$
 \left(1-\frac{n}{n^2} \right) \left(1-\frac{n}{n^2-1} \right)  \cdot \ldots \cdot \left(1-\frac{n}{n^2-n+1} \right)\\
 = \underbrace{\left(1-\frac{1}{n} \right)^n}_{\to \text{e}^{-1}} \cdot \underbrace{\frac{1-\frac{n}{n^2-1}}{1-\frac{1}{n}}}_{\rightarrow 1}  \cdot \ldots \cdot \underbrace{\frac{1-\frac{n}{n^2-n+1}}{1-\frac{1}{n}}}_{\rightarrow 1}
$$
which I thought would solve my problem. But after a second look I see that splitting the limits in the product cannot be allowed. This would be the same nonsense as 
$$
\underbrace{\left(1-\frac{n}{n^2} \right)}_{\rightarrow 1} \underbrace{\left(1-\frac{n}{n^2-1} \right)}_{\rightarrow 1}  \cdot \ldots \cdot \underbrace{\left(1-\frac{n}{n^2-n+1} \right)}_{\rightarrow 1}.$$
Maybe anybody can make the situation clear to me.

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.
