Robert Gerbicz wrote: It is impossible. Indirectly suppose that it is possible to rearrange the digits of $ {2}^{n}$ to give another larger power of two, we can assume that this is $ {2}^{n+c}$ where $ 0<c<4$, because $ {2}^{n+4}>10*{2}^{n}$ so this has got more digits than $ {2}^{n}$. But if we rearrange the digits of a number then it is giving the same remainder modulo 9, because this is the sum of the digits of the number modulo 9. So here $ {2}^{n}\equiv{2}^{n+c}\; modulo \; 9$, from this $ {2}^{n}*({2}^{c}-1)\equiv 0 \; modulo \; 9$ and this is impossible, because $ {2}^{n}$ isn't divisible by 3 and $ {2}^{c}-1$ isn't divisible by 9.