Proof of $10^{n+1} -9n -10 \equiv 0 \pmod {81}$ I am trying to prove that $10^{n+1} -9n -10 \equiv 0 \pmod {81}$. I think that decomposing into 9 and then 9 again is the way to go, but I just cannot get there. Any help is greatly appreciated. 
\emph{edit} I originally posted this a $9^n$ not $9n$. Apologies. 

$$
\begin{align}
&10^{n+1}-9n-10\\[9pt]
&=9\left(10\frac{10^n-1}{10-1}-n\right)\\
&=9\left(10\left(10^{n-1}+10^{n-2}+\dots+10+1\right)+9n-10n\right)\\
&=9\left(10\cdot9\left(\frac{10^{n-1}-1}{10-1}+\frac{10^{n-2}-1}{10-1}+\dots+\frac{10-1}{10-1}+\frac{1-1}{10-1}\right)+9n\right)\\
&=81\left(10\left(\frac{10^{n-1}-1}{10-1}+\frac{10^{n-2}-1}{10-1}+\dots+\frac{10-1}{10-1}+\frac{1-1}{10-1}\right)+n\right)
\end{align}
$$
