<s>[INST] 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?
 [/INST] This might get you started:
$$\begin{align}
&\left(1-{1^2\over n}\right)+\left(1-{1\over n}\right)\left(1-{2^2\over n}\right)+\left(1-{1\over n}\right)\left(1-{2\over n}\right)\left(1-{3^2\over n}\right)+\cdots\\
&\quad=\left(1-{1\over n}\right)\left(1+\left(1-{2^2\over n}\right)+\left(1-{2\over n}\right)\left(1-{3^2\over n}\right)+\cdots \right)\\
&\quad=\left(1-{1\over n}\right)\left(\left(2-{2^2\over n}\right)+\left(1-{2\over n}\right)\left(1-{3^2\over n}\right)+\cdots \right)\\
&\quad=\left(1-{1\over n}\right)\left(1-{2\over n}\right)\left(2+\left(1-{3^2\over n}\right)+\left(1-{3\over n}\right)\left(1-{4^2\over n}\right)+\cdots \right)\\
&\quad=\left(1-{1\over n}\right)\left(1-{2\over n}\right)\left(\left(3-{3^2\over n}\right)+\left(1-{3\over n}\right)\left(1-{4^2\over n}\right)+\cdots \right)\\
&\quad=\left(1-{1\over n}\right)\left(1-{2\over n}\right)\left(1-{3\over n}\right)\left(3+\left(1-{4^2\over n}\right)+\left(1-{4\over n}\right)\left(1-{5^2\over n}\right)+\cdots \right)\\
\end{align}$$
</s>