We need to prove:
$$P[N(B)=0|N(A)=n]=\left(1-\frac{|B|}{|A|}\right)^n$$
The attempt:
Let $\bar{B}=A\text\B$
\begin{align}P[N(B)=0|N(A)=n]&=\frac{P[N(B)=0\bigcap N(A)=n]}{P[N(A)=n]} \\\\ & = \frac{P[N(B)=0\bigcap N(A\text\B)=n]}{P[N(A)=n]} \\\\ &=\frac{P[N(B)=0\bigcap N(\bar{B})=n]}{P[N(A)=n]}\\\\ &= \frac{\text {exp}[-\lambda|B|]\text{exp}[-\lambda|\bar{B}|\frac{\lambda|\bar{B}|^n}{n!}}{\frac{\lambda |A|^n}{n!}\text {exp}[-\lambda|A|]}\\\\ &=\text{exp}[-\lambda \left( |B|+|\bar{B}|-|A|\right)]\left(\frac{|\bar{B}|}{|A|}\right)^n \end{align}
At this point I would like to say that $\bar{B}=A-B$. Then we would have $|\bar{B}|=|A-B|$. Then I would like to also say that $|A-B|=|A|-|B|$, and I would have:
$$P[N(B)=0|N(A)=n]=\text{exp}[0]\left(\frac{|A|-|B|}{|A|}\right)^n$$
and the result follows, but can I actually say all of that (the text in bold)?