Let \(x\) be the maximum number on the board. Assume there are infinitely many non-prime \(t>x\) on the board. Its prime divisors must be less than \(x\) otherwise primes in between \(x\) and \(t\) (which are on the board) would divide \(t\). Let \(\alpha_{t_1},\alpha_{t_2},...,\alpha_{t_m}\) be the powers of the primes in prime factorization \(t\) greater than \(x\). There is at least one \(j\) such that \(\alpha_{k_j}\) does not have upper bound otherwise elements will be repeated. Let this \(j\) be 1. Thus we have a infinite subsequence in which \(\alpha_{k_1}\) of the elements of the sequence is increasing. We can apply this algorithm again for some other exponent because \((\alpha_{t_2},...,\alpha_{t_m})\) cannot have repeated terms. At the end of this algorithm we will have a infinite sequence and this sequence have infinitely many elements whose exponents are in increasing order, a contradiction.

Let $M$ be the largest number of the initial set, and $S$ be the sequence of numbers. Then, note that for all primes $p>M$ we clearly have $p\in S$. Secondly, note that for all primes $p$ there exists a unique positive integer $k$ such that $p^k\in S$. There is at least one because consider the smallest $K$ such that $p^{K}>M$. Either $p^K\in S$, or there is some $s\mid p^K$, so $s=p^k\in S$ and we're done. The uniqueness is clearly true (and also not very useful). Let $f(p)$ be the power of the largest prime power that is $\leq M$ and not in $S$. Thus, any composite number $n\in S$ satisfies \[v_p(n) \leq f(p)+1\]However, for $q>M$, $q\nmid n$, so if $P$ is the set of primes $<M$, then any composite $n\in S$ satisfies \[n\leq \prod_{p\in P} p^{f(p)+1}\]Where the RHS is clearly finite. Thus, since all primes appear, eventually the only numbers being written are primes.

Let $A$ be set of numbers $a_1 < a_2 < ... < a_n$ on board. Let $B$ be set of prime numbers $p_1 < ... < p_m$ such that $p_m < a_n$ and and are not on board. Note that for each number Nordi writes we have $2$ cases prime or composite. Note that any prime number after $a_n$ will appear on board cause fits the conditions. Now for number $X$ which Nordi writes, we can't have $p_i \mid X$ and $p_i $ not in $B$ cause then $p_i$ is somewhere on board. so we must choose prime factors of $X$ from $B$ which can be done finitely so from some point no such $X$ can be written so $X$ can only be prime from that point on.