Prove that a number that consists of $3^n$ ones is divisible by $3^n$ I can't even make sense of this question. Isn't this just like asking, "Prove that 3 is divisible by 3." Isn't any number divisible by itself? Is this all there is to this question—it seems like there must be more to it.

We can use induction. I prefer to show that the number that "consists of" $3^n$ $9$'s is divisible by $9\cdot 3^n$.
The number whose decimal representation consists of $3^n$ consecutive $9$'s  is $10^{3^n}-1$. 
For the induction step, note that $10^{3^{k+1}}-1=x^3-1$ where $x=10^{3^k}$. This factors as $(x-1)(x^2+x+1)$. By the induction assumption, $x-1$ is divisible by $9\cdot 3^k$. Also, $3$ divides $x^2+x+1$, so $9\cdot 3^{k+1}$ divides $x^3-1$.
Remark: Or else we could show that the number whose decimal representation consists of $3^{k+1}$ consecutive $1$'s is the number with $3^k$ consecutive $1$'s, times a number of the shape $1000\cdots 01000\cdots 01$. The second number is divisible by $3$ by the usual divisibility test.
