Since $A$ is a nonempty subset of the positive integers, $A$ has a minimum element $m$. If $m > 1$ then $m > \sqrt[3]{m} \ge [\sqrt[3]{m}]$ and $[\sqrt[3]{m}] \in A$. It is contrary to that $m$ is the minimum element. So $m = 1$.

Since $1 \in A$, $9^k \in A$. From this $[\sqrt[3]{81}] = 4 \in A$ and $4 \cdot 9 = 36 \in A$ and $[\sqrt[3]{36}] = 3 \in A$. Then $3^n \in A$ for $n = 1, 2, \dots$ (*).

Lemma. There exists $3^k$ type integer in the $[n, 3n]$ intervalum.

Proof of lemma. Let $3^s \le n < 3^{s+1}$. Then $n < 3^{s+1} \le 3n$. $\square$

Assume that there exists $n$ such that $n \notin A$. Let us show that if $a \in [n^{3^p}, (n+1)^{3^p} - 1]$ then $a \notin A$. Suppose that $a \in A$, then $[\sqrt[3]{a}] \in A$ and $[\sqrt[3]{a}] \in [n^{3^{p-1}}, (n+1)^{3^{p-1}} - 1]$. Using this statement $p$ times, we'll get $[\sqrt[3]{\dots[\sqrt[3]{[\sqrt[3]{a}]}]} = n \in A$ which is contradiction.

Since $\log_3(n+1) - \log_3 n > 0$ and $\lim_{k \to \infty} \frac{1}{3^k} = 0$, there exists a positive integer $k$ such that
$$\log_3(n+1) - \log_3 n > \frac{1}{3^k}.$$
From this $3^k \log_3 \frac{n+1}{n} > 1$, then $\log_3 \left(\frac{n+1}{n}\right)^{3^k} > \log_3 3$ and $(n+1)^{3^k} > 3 \cdot n^{3^k}$. Hence $[n^{3^k}, 3n^{3^k}] \subseteq [n^{3^k}, (n+1)^{3^k} - 1]$. By the lemma, there exists $s \in \mathbb{N}$ such that $3^s \in [n^{3^k}, (n+1)^{3^k} - 1]$ and $3^s \notin A$, contradicting (*). It means $\mathbb{N} \subseteq A$.