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. 

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.
