Starting from the length-1 list whose only entry is 1, iterate the process of replacing the last (and largest) entry in the list of length $n$ (call that entry $m$) by the two numbers $mU_n$ and $m(1-U_n)$, where $U_1,U_2,\dots$ are independent identical draws from the uniform distribution on $[0,1]$, and sorting the new list from smallest to largest. Show that after $n-1$ steps the length-$n$ list that we see is likely to be close to $(1/2n^2,2/2n^2,\dots,n/2n^2)$.

This must be a classic fragmentation process, but five minutes of Googling failed to turn up anything relevant.