Showing that a $3^n$ digit number whose digits are all equal is divisible by $3^n$ 
Let $c$ be a $3^n$ digit number whose digits are all equal. Show that $3^n$ divides
  $c$.

I have no idea how to solve these types of problems. Can anybody help me please?

A number whose digits are all equal and of length $3^n$ is thus of the form $c = \sum\limits_{i=0}^{3^n-1} a \cdot 10^i = a \dfrac {10^{3^n}-1}{10-1}$ by the geometric series.
Since we have to account for the possibility that $a = 1$ we need to show that $3^n \mid \dfrac{10^{3^n}-1}9$, i.e. $3^{n+2} \mid 10^{3^n}-1$.
This we can do by induction: Base case is trivial, suppose $3^{n+2} \mid 10^{3^n}-1$.
Then we have $10^{3^n} \equiv 1 \pmod {3^{n+2}}$, and so: $10^{3^n} \equiv 1 + k 3^{n+2} \pmod {3^{n+3}}$. Now compute $10^{3^{n+1}} = \left(10^{3^n}\right)^3$ modulo $3^{n+3}$.

Addendum: In general, when considering repdigits it is useful to have them in the form obtained through the geometric series. Since the length of the repdigit is in the exponent, and exponents are well-behaved under modulo calculations, it is one of the most informative forms for discovering properties of repdigits; in particular those pertaining to divisibility.
