Proving a number with $3^n$ equal digits is divisible by $3^n$ Prove a number with $3^n$ equal digits is divisible by $3^n$.
My thoughts about  the problem are: a number with $3^n$ equal digits $d$ is equal to $d\frac {10^{3^n} - 1} {9}$. We use Lifting The Exponent lemma, or plain induction.

This follows from the fact that $x^3-1=(x-1)(x^2+x+1)$ and that $10\equiv1\pmod{3}$. That is,
$$
10^{3^n}-1=\left(10^{3^{n-1}}-1\right)\left(10^{2\cdot3^{n-1}}+10^{3^{n-1}}+1\right)\tag{1}
$$
and
$$
\begin{align}
\left(10^{2\cdot3^{n-1}}+10^{3^{n-1}}+1\right)
&\equiv1+1+1\\
&\equiv0\pmod{3}\tag{2}
\end{align}
$$
so that
$$
\left.3\,\middle|\,\left(10^{2\cdot3^{n-1}}+10^{3^{n-1}}+1\right)\right.\tag{3}
$$
We start out with
$$
\left.3^2\,\middle|\,\left(10^{3^0}-1\right)\right.\tag{4}
$$
Next, combining $(1)$ and $(3)$ yields that
$$
\left.3^{n+1}\,\middle|\,\left(10^{3^{n-1}}-1\right)\right.\Rightarrow\left.3^{n+2}\,\middle|\,\left(10^{3^n}-1\right)\right.\tag{5}
$$
Therefore, induction on $n$, using $(4)$ and $(5)$, says
$$
\left.3^n\,\middle|\,\frac{10^{3^n}-1}{9}\right.\tag{6}
$$
