"$111 \dots$ upto $3^n$ digits" is divisible by $3^n$ 
Prove that an integer of the form "$111 \dots$ upto $3^n$ digits" is divisible by $3^n$

My attempt
For $n=1,$ $111$ is divisible by 3.
Let $T_n=111...$ upto $3^n$ digits is divisible by $3^n$.
Then $T_{n+1}=111...$ upto $3^{n+1}$ digits. I am unable to prove that Then $T_{n+1}$ is divisible by $3^n$.

step 1. Write the number K = 111... in decimal form as 1*100 +  1*101 +  1*102 +... = $\sum_{i=0}^m10^i$
step 2. Examine K mod $3^n$ = $\sum_{i=0}^m10^i$ mod $3^n$ = $\frac{1-10^{m+1}}{1-10}$ mod $3^n$ = $\frac{10^{m+1}-1}{9}$ mod $3^n$ = $\frac{10^{3^n+1} -1}{9}$ mod $3^n$. Where the first equality follows from the formula for the sum of a finite geometric series. Clearly, 9|$10^{3^n+1} -1$ is necessary, so compute K' = $10^{3^n+1} -1$ mod (9*$3^n$)= $10^{3^n+1} -1$ mod $3^{n+2}$. If K' = 0, we're done.
step 3.
$\phi$($3^{n+2}$) =  $3^{n+1}$-$3^n$ = a
Thus, by Euler's formula, $10^a$ = 1 mod $3^{n+2}$. Therefore, K' + 1 = $(10^a)^{3^n}$= 1 mod $3^{n+2}$.
So, $10^{3^n +1}$ -1 mod $3^{n+2}$ = 0. 
The End
