In the 1950s, Sierpiński asked whether there exists a linear order that is isomorphic to its lexicographically ordered Cartesian cube but not to its square. The analogous question has been answered positively for many different classes of structures, including groups, Boolean algebras, topological spaces, graphs, partial orders, and Banach spaces. However, the answer to Sierpinski's question turns out to be negative: any linear order that is isomorphic to its cube is already isomorphic to its square, and thus to all of its finite powers. I will present an outline of the proof and give some related results.