Independent Event Product Rule Proof

1. Suppose $E$ and $F$ are Independent Events
2. Suppose $P(E)>0$
3. Suppose $P(F)>0$
4. Then, $P(E|F)=\frac{P(E\cap F)}{P(F)}$ by Conditional Probability
5. Rearrange to get $P(E\cap F)=P(E|F)\ast P(F)$
6. Note that $P(E|F)=P(E)$ by 1, as $E$ and $F$ are Independent Events
7. Thus $P(E\cap F)=P(E)\ast P(F)$ $\square$