Prove that every positive integer having $3^m$ equal digits is divisible by $3^m$ 
Prove that every positive integer having $3^m$ equal digits is divisible by $3^m$  

I haven't really done anything substantial to solve this but here's what I have:
Let the positive integer be $aaa....aaa$ or simply $A$.    
$A=a\cdot10^{3^m}+a\cdot10^{3^m-1}+a\cdot10^{3^m-2}+....+a$
This forms a GP, hence, 
$A=x\frac{10^{3^m+1}-1}{10-1}=\frac{x}{9}\ [10^{3^m+1}-1]$   
Now the question wants us to prove
$3^m|A$
$3^m|\frac{x}{9}\ [10^{3^m+1}-1]$
$3^{m+2}|\ x (10^{3^m+1}-1)$
But x can be any digit so,
$3^{m+2}|\ 10^{3^m+1}-1$  
I can't figure out a way to solve further. Can I get a hint as to how I can proceed with this question
I have tried induction too, but didn't really reach the conclusion:   
Let P(n) be
$10^{3^n+1}-1=3^{n+2}a$    
P(1):
$10^{3+1}-1$
$=10^4-1$
$=10,000-1$
$=9999$, which is divisible by $3^3=27$
Let p(m) be true,
$10^{3^m+1}-1=3^{m+2}b$  
To prove: P(m+1) is also true
$10^{3^{m+1}+1}-1$
$=10^{3*3^m+1}-1$
$=10^{2*3^m}\cdot10^{3^m+1}-1$
$=10^{2*3^m}(3^{m+2}b+1)-1$    
I am not able to go further...
 A: We will proceed by induction on $m$ starting from $m=0$. Let $d$ be any digit from 1 to 9. We need to prove this statement for all $m\ge0$:
$$S(m): 3^m\mid\underbrace{ddd\dots ddd}_{3^m}$$
Clearly $S(0)$ is true:
$$3^0\mid d\to1\mid d$$
Now assume that $S(n)$ for some $n$ is true.
$$3^n\mid\underbrace{ddd\dots ddd}_{3^n}$$
The concatenation of three copies of $\underbrace{ddd\dots ddd}_{3^n}$ can be represented as a product:
$$\underbrace{ddd\dots ddd}_{3^{n+1}}=\underbrace{ddd\dots ddd}_{3^n}×(10^{2\cdot3^n}+10^{3^n}+1)$$
The term $\underbrace{ddd\dots ddd}_{3^n}$ is divisible by $3^n$, while $(10^{2\cdot3^n}+10^{3^n}+1)$ is divisible by 3:
$$\begin{align}
10^{2\cdot3^n}+10^{3^n}+1&\equiv
1^{2\cdot3^n}+1^{3^n}+1\bmod3\\
&\equiv1+1+1\bmod3\\
&\equiv0\bmod3
\end{align}$$
Putting these terms together, we see that
$$(3×3^n)\mid\underbrace{ddd\dots ddd}_{3^{n+1}}$$
$$3^{n+1}\mid\underbrace{ddd\dots ddd}_{3^{n+1}}$$
but this is precisely the statement $S(n+1)$. Thus if $S(n)$ is true then $S(n+1)$ is true. Since $S(0)$ is true, by induction $S(m)$ is true for all $m\ge0$.
A: Your idea for an inductive proof can be made to work as follows: 
$\begin{eqnarray} 
3^m\mid 11\cdots 11 = (10^{\large 3^{\Large m}}\!\!-1)/9\iff&\! 10^{\large 3^{\Large m}} =&\ 1 + 3^{\large m+2} k,\,\ \ \ {\rm i.e.}\ \ P(m)\\ \Rightarrow\ &  10^{\large 3^{\Large m+1}}\!\! =& (1 + 3^{\large m+2} k)^{\large 3}\\
 &\qquad\  =&\ 1+ 3^{\large m+3}\, k',  \ \ {\rm i.e.}\ \ P(m\!+\!1) \end{eqnarray}$
Remark $\ $ This is a special case of LTE = Lifting The Exponent Lemma
