Consider $n$ independent trials, each of which results in one of the outcomes $1,\ldots,k$ with respective probabilities $p_1,\ldots,p_k$, $\sum_{i=1}^{k} p_i =1$. (I interpret this summation as just saying mathematically that definitely one of the outcomes has to occur on each trial). Show that if all the $p_i $are small, then the probability that no trial outcome occurs more than once is a approximately equal to $ \exp\left(−n(n−1)\sum_i \frac{p_i^2}{2}\right). $
So if all the $p_i$are small, in comparison to $n$, then I believe I can approximate this to a Poisson RV (that is where the $\exp$ comes in). (Correct?) So, first I can write the prob that no trial occurs more than once as$ p_1(1−p_1)^{n−1} p_2 (1−p_2)^{n−1}\cdots p_k(1−p_k)^{n−1} \cdot k!=k!\prod_{i=1}^{k} p_i(1−p_i)^{n−1}$ I am not really sure where to go from here. I am guessing at some stage, I will need to take the limit that as $n$ tends to infinity, something tends to $e $ .Thanks.