1. Assume for contradiction: Suppose the problem statement is false. Let $t$ be the smallest positive integer that does not appear in the sequence. Therefore, there must exist $N \in \mathbb{Z}_{>0}$ such that $\{1, 2, \dots, t-1\} \subset \{a_0, a_1, a_2, \dots, a_N\}$.

2. Define k-good numbers: A positive integer $l$ is called *k-good* if:
- $l \equiv 0 \pmod{p_1 p_2 \dots p_{k-1}}$
- $l \equiv 1 \pmod{p_k}$
where $p_i$ is the $i$-th prime number.

3. Observation on k-good numbers: Note that $a_n$ cannot be a k-good number for any $n > N$; otherwise, we would have $a_{n+1} = t$ (since $p_k = p(a_n) \mid (a_n^t - 1)$ and $1, 2, \dots, t-1$ have already appeared), which contradicts our assumption. Therefore, there are finitely many k-good numbers that appear in the sequence.

4. **Lemma on $p(n)$**: We need to show that $p(n) < D \log n$ for some constant $D$ and for all $n$.

- Let $p_k = p(n)$. By definition, $p_1 p_2 \dots p_{k-1} \mid n$, so $k! < p_1 p_2 \dots p_{k-1} \le n$.
- Using the Prime Number Theorem and Stirling's approximation, we know that $k > C_1 \frac{p_k}{\log p_k}$ and $k! > C_2 \left(\frac{k}{e}\right)^k$ for some constants $C_1, C_2$.
- Therefore, $p(n) < D \log n$ for some constant $D$.

5. Choose large constants: Take sufficiently large constants $P, Q$. Since the sequence $\{a_n\}_{n \ge 0}$ passes through each positive integer at most once, there must exist an $n$ satisfying:
- $n > P$
- $a_n > Q$
- $a_{n+1} > a_n$

6. **Consider $q = p(a_n)$**: Let $L$ be a positive integer larger than the number of k-good numbers in the sequence. Let $k$ be such that $p_k = p(q-1)$.

- By the choice of $k$ and the lemma, we have:
- $p_k \nmid q-1$
- $L \cdot \text{lcm}(p_1, p_2, \dots, p_k, q-1) = L \cdot p_k \cdot (q-1) < a_n$
- This inequality holds because $q < D \log(a_n)$ and $p_k < \log(q-1) < D \log(\log a_n)$, and so the inequality must hold when taking $a_n$ large enough.

7. Apply the Chinese Remainder Theorem: The two conditions above imply that there must exist at least $L$ solutions smaller than $a_n$ to:
- $x \equiv 0 \pmod{\text{lcm}(p_1, p_2, \dots, p_{k-1}, q-1)}$
- $x \equiv 1 \pmod{p_k}$

Since all of these numbers are k-good and there are fewer than $L$ k-good numbers in the sequence, one of these numbers, say $c$, has not appeared in the sequence and satisfies all the above conditions. Note that since $q-1 \mid c$, we must have $q \mid a_n^c - 1$, and so $a_{n+1} \le c \le a_n$, which contradicts the choice of $n$.

8. Conclusion: Therefore, our initial assumption that $t$ does not appear in the sequence is false. Hence, every positive integer appears as a term in the sequence.

$\blacksquare$