Proving that L is not regular by showing that $\equiv_L$ has infinite index

Question

Proving that L is not regular by showing that $\equiv_L$ has infinite index.
$\Sigma$ = {a}, L = {$a^{3^n} : n \geq$ 0}
My ideas:
theorem of Myhill-Nerode: L $\in$REG $\Leftrightarrow$ $\equiv_L$ has finite index
We show: $\equiv_L$ has infinte equivalence classes.
$a^{3^i}$ $\not\equiv$ $a^{3^{i+k}}$, $k>0$
-> equivalence classes: [a], [aaa], [$a^3$]...
Every $a^{3^i}$ is a different equivalence class. Therefore $\equiv_L$ has an infinite amount of equivalence classes $\rightarrow$ L is not regular. qed.

Answer 1

We will show that $a^{3^t}$ is not equivalent to $a^{3^{t+k}}$ for any $k>0$. Two words $x,y$ are Myhill-Nerode equivalent iff for every word $z\in\Sigma^*$, either both $xz$ and $yz$ are in the language or neither $xz$ nor $yz$ are in the language. Let $x=a^{3^t}$, $y=a^{3^{t+k}}$, $z=a^{(2*3^t)}$. $$ xz = a^{3^t}a^{(2*3^t)}=a^{3^{t+1}} \\ yz = a^{3^{t+k}}a^{(2*3^t)}=a^{3^t(3^k+2)} $$ Clearly $xz$ is in the language. However, $yz$ is not in the language, since $3^t(3^k+2)$ is not a power of $3$.

Therefore no two element of $L$ are in the same equivalence, so there are an infinite number of equivalence classes.

Answer 2

Hint. In order to prove that for $k > 0$, $a^{3^i} \not\equiv a^{3^{i+k}}$, just go back to the definition of the Nerode equivalence. You need to prove that there exists a word $u_k = a^{n_k}$ such that $a^{3^i}a^{n_k} \in L$ but $a^{3^{i+k}}a^{n_k} \notin L$ or $a^{3^i}a^{n_k} \notin L$ but $a^{3^{i+k}}a^{n_k} \in L$. Can you find such a word for every $k$?