Find all $n\in\mathbb{Z}^+$ such that the sum of the digits of $5^n$ equals $2^n$
Starting with a table of values, I found that $n=3$ works. Beyond this, it's hard to imagine any other number working. As $n$ gets larger, it seems to me that $2^n$ grows far more rapidly than the sum of the digits of $5^n$ would ever be able to "catch up" to.
My reasoning: multiplying by each $2$ doubles $2^n$, but multiplying by each $5$ adds (I think) at most $1$ digit to $5^n$, which would add at most $9$ to the sum of the digits of $5^n$. Another way of looking at it, suppose $n=1000000$. Notice that $5^{1000000}<10^{1000000}$, which is $1$ with $1000000$ zeros. Thus $5^{1000000}$ has at most $1000000$ digits, and the sum of its digits would be maximized if those digits were all $9$s. Then the sum of the digits of $5^n$ would be $9\cdot1000000=9000000$, which is far less than the number $2^{1000000}$.
But is there a way to prove this using theory?