Why is this sum zero? I have been looking at the following sum (for any positive integer $n$)
$$\left(1-\frac{1^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3^2}{n}\right) + \ldots $$
Note that the $i$th term in the sum has $i$ factors and  is 
$$\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3}{n}\right)\dots \left(1-\frac{i-1}{n}\right)\left(1-\frac{i^2}{n}\right).$$  
It seems, amazingly, that the answer is 0. How can one show this?

The partial sums are 
$$\begin{eqnarray}
1/2&0\\
2/3&4/9&0\\
3/4&12/16&18/64&0\\
4/5&24/25&72/125&96/625&0\\
5/6&40/36&180/216&480/1296&600/6^5&0\end{eqnarray}$$
The ratios from one partial sum to the next are 
$$\begin{eqnarray}0\\
2/3&0\\
4/4&(3/2)/4&0\\
6/5&(6/2)/5&(4/3)/5&0\\
8/6&(9/2)/6&(8/3)/6&(5/4)/6&0\end{eqnarray}$$
So is the sum of the first $k$ terms, with $n$ as the key variable, 
$$\frac{\frac{(n-1)!}{(n-1-k)!}k}{n^k}$$
