Prove that there is no integer power of $2$ that is the result of a reordering of the digits of another power of $2$.
Attempt: Say we are working with $2^k$, with $n$ digits.
If we reordered it and got another power of $2$, then the new number must be of $2^{k-3}, 2^{k-2}, 2^{k-1}, 2^{k+1}, 2^{k+2} , 2^{k+3}$.
However none of these numbers are congruent $2^k \text{mod 9}$, we've reached a contradiction.
Unless if we could put the zero in front to shorten the length of the original number, which is still a case I am working on...
Attemp 2: I thought in supposing, in order to obtain absurdity, that there is indeed an integer power of 2 which is the result of a reordering of the digits of another power of 2.
Let $2 ^ m$ be the power equal to the reordering of the digits of power $2 ^ n$; for $m, n$ nonnegative integers.
Let's suppose that $m$; that is, we do not consider application identity as a possible reordering. We can then consider m and n positive, since the only solution with one of these nulls would be $m = n = 0$, for the power $2 ^ 0 = 1$.