 Independent exponential order-statistic spacings

For $n$ independent and identically distributed rate-$\lambda$ exponential random variables, let $M_0=0$ and $M_1<\cdots <M_n$ be their order statistics. The gaps $D_i=M_i-M_{i-1}$ are independent and have rates $(n-i+1)\lambda$. After the first clock rings, the memoryless property makes the surviving clocks fresh independent exponentials, independently of the elapsed minimum and the winning label. Induction gives the result. Equivalently, the ordered joint probability density is $n!{\lambda }^n\exp [-\lambda {\sum }_i{m}_i]$ on $0<m_1<\cdots <m_n$; changing to gaps gives a product of exponential densities because ${\sum }_i{m}_i={\sum }_i(n-i+1)D_i$.