Digit sum of $9^n$

Let $n>2$ be an integer. Prove that the digit sum of $9^n$ is greater than $9$.

My starting approach was to observe that $9^n$ either ends in $1$ or $9$. If it ends in $9$ then the result is obvious. If it ends in $1$, I'm not sure what to do.

Solution 1:

It looks like this can be shown by generalising the argument in the question and checking lots of cases. Assume for contradiction that the digit sum of $9^n$ is $9$. Then either $9^n < 100000$, or the following three properties hold.

• The digit sum of $9^n \% 100000$ is less than 9
• The digit sum of $9^n \% 99999$ is exactly 9
• The last digit of $9^n \% 99999$ is not zero

where $9^n\%d$ means the least non-negative residue of $9^n$ modulo $d$.

This Python code checks $n \leq 7501$, after which the pairs $(n\%100000, n\%99999)$ start repeating with period 7500. It shows that there are no $n$ for which the three properties all hold.