Solution. Denote $b_n = \frac{a_n}{\text{rad}(a_n)}$. Since rad($a_n$) divides rad($a_{n+1}$) we have $b_{n+1} | b_n + 1$. If there are indices $i < j$ with $b_i < 2022 < b_{i+1}$, we will be done by “continuity”. If, to the contrary, this does not happen, there are two possible cases.
* $b_n < 2022$ for all $n$ big enough. Since $a_n$ increases indefinitely, then so does rad($a_n$), so at some moment rad($a_n$) receives a new prime $p > 2022$. This means that $p \nmid a_n$ and $p | a_{n+1} = \frac{b_{n+1}}{b_n}a_n$, so $p | b_n + 1$ and hence $b_n \ge 2022$, a contradiction.
* $b_n > 2022$ for all $n$. We can assume WLOG that $b_0$ is the smallest term of the sequence $(b_n)$. Suppose that $b_{i+1} = b_i + 1$ for all $0 \le i < n$. Then
$$\text{rad}(a_0) = \dots = \text{rad}(a_{n-1}) = R.$$
But for every prime $p \le n$ there is a multiple of $p$ among us sus $b_0, \dots, b_{n-1}$, so $p | a_k$ for some $k$ and consequently $p | R$. Since not every prime divides $R$, there must be an index $n$ such that $b_n < b_{n-1} + 1$, i.e. $b_{n-1} + 1 = db_n$ for some $d > 1$ and $\text{rad}(a_{n+1}) = dR$, so $\text{gcd}(d, R) = 1$.
Recall that $b_0 \le b_n = \frac{b_0+n}{d}$, which reduces to $n \ge (d-1)b_0$. By above, this means that all primes up to $(d-1)b_0$ divide $R$, but $d$ does not divide $R$, so $d > (d-1)b_0$, which is impossible. $\square$