Let $M$ be a measurable space with a measure $\mu$. When $M$ is finite or countable, $N$ is a Poisson point process of intensity $\mu$, and when $A$ is a measurable set such that $\mu(A)$ is finite, then conditionally on $N$ we can uniformly choose an order of the $N(A)$ elements of the process in $A$ among all possible $N(A)!$ choices. In this way we obtain a finite ordered random family $U_1, \dots , U_{N(A)}$ and we see that $$ \mathbb P(U_1 = u_1, \dots , U_n = u_n\lvert N(A) = n)= \frac{\mu(u_1)\dots\mu(u_n)}{\mu(A)^n}$$ and $$ \mathbb P(U_1 = u_1, \dots , U_n = u_n, N(A) = n)= \frac{\mu(u_1)\dots\mu(u_n)}{n!}e^{-\mu(A)}$$
I understand how they got the second equality thanks to the first one but I don't get where the first equality comes from.
Let $B = A\backslash\left\{u_1,\ldots,u_n\right\}$
\begin{align} \mathbb P\left[U_1 = u_1,\ldots,U_n=u_n,N(A) = n\right] &= \frac1{n!}\mathbb P\left[\bigcap_{i=1}^n\left\{N\left(\left\{u_i\right\}\right)= 1\right\}\cap \left\{N(B) = 0\right\}\right]\\ &=\frac1{n!}\prod_{i=1}^n \frac{\mu(u_i)}{1!}e^{-\mu(u_i)}\times e^{-\mu(B)}\\ &= \frac{\mu(u_1)\cdots\mu(u_n)}{n!} e^{-\sum_{i=1}^n \mu(u_i) - \mu(B)} = \ldots \end{align}