Every linear order isomorphic to its cube is isomorphic to its square
  In 1958, Sierpinski asked whether there exists a linear order $X$ that is isomorphic to its lexicographically ordered cube but is not isomorphic to its square. The main result of this paper is that the answer is negative. More generally, if $X$ is isomorphic to any one of its finite powers $X^n$, $n>1$, it is isomorphic to all of them. 
Garrett Ervin
math.LO