How to show that $10^n - 1$ is divisible by $9$ How can I show that $10^n-1, 10^{n-1}-1,...., 10-1$ are all divisible by 9? I was considering using Euclid's algorithm, but I can't find a way to get that to work.

$$ a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) $$
Proof of the identity by expansion: 
$$\begin{align}
&(a-b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) \\\\
&=a^{n} + a^{n-1}b + a^{n-2}b^2 + \ldots + ab^{n-1}\\\\
&\;\;\;\;\;\;\;\;- ba^{n-1} - b^2a^{n-2}-\ldots -b^{n-1}a - b^n\\\\
&=a^n-b^n
\end{align}$$
