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. 

Without anything more thatn the geometric sum formula:
It's certainly true mod $9$, since it is easy to reduce the left side by $9$: 
$$1^{n+1}-0-1\equiv{0}\mod{9}$$ 
Now dividing the left side by $9$ gives $$10\frac{10^n-1}{9}-n$$ which is $$10\frac{(10^{n-1}+10^{n-2}+\cdots+10^1+1)(10-1)}{9}-n$$ which is $$10(10^{n-1}+10^{n-2}+\cdots+10^1+1)-n$$ Reducing mod $9$ gives $$1\cdot\left(\overbrace{1+1+\cdots+1}^{n\text{ copies}}\right)-n\equiv0\mod{9}$$
So the original left hand side is divisible by $9$, twice. Thus it's divisible by $81$.
