Proof for the uniqueness of all the combination of the digits of $2^k$, where k>3

Question

A long time ago I found a question on the internet that went a little like this:

Suppose that we have $n=2^k$ where $k\gt 3$. If $m$ is another number that is a combination of the digits of $2^k$, prove that $m$ cannot be a power of $2$.

I gave up on it a long time ago, but have now become interested in number theory and hope that someone could shed some light on this problem.

Edit: This is only for base 10.

Answer 1

Assuming you mean shuffle the digits around (in base $10$)

For the case where you don't bring zeroes to the front:

We look at the remainder when you divide by $9$.

If $2^m = 2^n$ modulo $9$, then $m-n$ is divisible by $6$. In which case, they have different number of digits.