If all the numbers $(1^\alpha,\,2^\alpha,\,3^\alpha,\,\dotsc)$ are integer, then $\alpha$ is an integer. A theorem of Siegel asserts that


  If $\beta&gt;0$ and $2^\beta,\,3^\beta,\,5^\beta$ are integers, then $\beta$ is an integer.


The following result is a beautiful consequence of this theorem


  If $\beta$ is a positive number such that $1^\beta,\,2^\beta,\,3^\beta,\,\dotsc$ are integers, then $\beta$ is itself is an integer.


I'm looking for a proof of this result.

Note. This result appeared as a problem in the 1972 Putnam Prize competition, and not one of more than 2000 university student competitors gave a solution; the solution, though not hard, could well elude even a professional mathematician for several hours (or days).
Well, k1.M, here an extension of important and beautiful Siegel Theorem (I did not know). It is clear $\beta$ must be integer or irrational and by another theorem of Siegel $\beta$ must be trascendental. I guess the statement goes for any three coprime integer x, y, z the triple 2, 3, 5 having been chosen for smallest.Thus the true spirit of the statement of Siegel would be "all trascendental as exponent of an integer can gives an integer at most twice".
Well, k1.M, here an extension of important and beautiful Siegel Theorem (I did not know). It is clear $\beta$ must be integer or irrational and by another theorem of Siegel $\beta$ must be trascendental. I guess the statement goes for any three coprime integer x, y, z the triple 2, 3, 5 having been chosen for smallest.Thus the true spirit of the statement of Siegel would be "all trascendental as exponent of an integer can gives an integer at most twice".