What is the $\prod \frac{1}{n-1}$ I am trying to compute the probability of none the events occurs where the probability for each event is  $Pr[A_i]=\frac{1}{n-1}$ for all i and these events are independent. 
What is the $\prod_{i=3}^{n} \frac{1}{n-1}$ when n >= 3
I know that the Pr(none event occur) = 1 - Pr(at least one occur)
= 1 - $\prod_{i=3}^{n} \frac{1}{n-1}$
I want to proof that the probability that none of them occur is ≥1/8

*

*$\prod_{1}^{n} \frac{1}{n-1} = (\frac{1}{n-1})^{n}$

*$\Pr($none event occur$) = 1-(\frac{1}{n-1})^{n}$


For any $n\geq 3$, we have
$log(\frac{8}{7}) < n.log(n-1) \Rightarrow -log(\frac{8}{7}) > -n.log(n-1) \Rightarrow log(\frac{8}{7})^{-1} > n.log(n-1)^{-1} \Rightarrow log(\frac{7}{8}) > n.log(\frac{1}{n-1}) \Rightarrow \frac{7}{8} > (\frac{1}{n-1})^{n} \Rightarrow 1-\frac{1}{8}>(\frac{1}{n-1})^{n} \Rightarrow 1-(\frac{1}{n-1})^{n} >\frac{1}{8} \Rightarrow \Pr(none\text{ }event\text{ }occur) > \frac{1}{8}$
(All logarithms are in base 2)
