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.

May be another way to see the limit and how it is approached.
Let $$u_i=1-\frac 1{n^2-i}\implies P_n=\prod_{i=0}^{n-1} u_i$$ Take logarithms $$\log(P_n)=\sum_{i=0}^{n-1} \log(u_i)$$ Now, use Taylor series for large values of $n$ $$\log(u_i)=-\frac{1}{n}-\frac{1}{2
   n^2}-\frac{i+\frac{1}{3}}{n^3}-\frac{i+\frac{1}{4}}{n^4}+O\left(\frac{1}{n^5
   }\right)$$ Now, compute the sums $$\log(P_n)=-1-\frac{1}{2 n}+\frac{1-3 n}{6 n^2}+\frac{1-2 n}{4 n^3}+\cdots=-1-\frac 1 n-\frac 1 {3n^2}+\frac 1 {4n^3}+\cdots$$  Going back to the exponential $$P_n=\frac 1e\Big(1-\frac{1}{n}+\frac{1}{6 n^2}+O\left(\frac{1}{n^3}\right) \Big)$$
