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.

$10^{3^n}-1$
=$-1+(1+3^2)^{3^n}$
=$-1+ 1+(3^nC_1)(3^2)+(3^nC_2)(3^2)+\cdot\cdot\cdot+(3^2)^{3^n}$
which will be divisible by $3^{n+2}$ if n+4≥n+2 , (n-1+6)≥n+2, $\cdot\cdot\cdot\ , 2.3^n≥n+2$ which is true for n≥0.
So, $\frac{10^{3^n}-1}{9}$ will be divisible by $3^n$  for n≥0.
