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. 
 A: I would use induction. It works for $n=0$ Now subtract the $n$ case from the $n+1$ case.
A: We could use the binomial theorem:
$$10^{n+1} \equiv (1+9)^{n+1} \equiv 1+9\binom{n+1}{1}+9^2\sum_{i=2}^{n+1}{9^{i-2}\binom{n+1}{i}} \equiv 1+9(n+1)\pmod{81}$$
Thus $10^{n+1}-9n-10 \equiv 0 \pmod{81}$.
A: $$
\begin{align}
&10^{n+1}-9n-10\\[9pt]
&=9\left(10\frac{10^n-1}{10-1}-n\right)\\
&=9\left(10\left(10^{n-1}+10^{n-2}+\dots+10+1\right)+9n-10n\right)\\
&=9\left(10\cdot9\left(\frac{10^{n-1}-1}{10-1}+\frac{10^{n-2}-1}{10-1}+\dots+\frac{10-1}{10-1}+\frac{1-1}{10-1}\right)+9n\right)\\
&=81\left(10\left(\frac{10^{n-1}-1}{10-1}+\frac{10^{n-2}-1}{10-1}+\dots+\frac{10-1}{10-1}+\frac{1-1}{10-1}\right)+n\right)
\end{align}
$$
A: 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 $81$.
A: The original expression can be factored as $9(10^n+10^{n-1}+\cdots+1-n)$. The quantity in parenthesis consists of a number all of whose n digits are are ones, minus n. Such a number is divisible by 9, since any number, mod 9, is equal to the sum of the digits, mod 9.
