Since $a_{n+1}=a_n(a_n+1)$, it follows that $a_1|a_2|...|a_n$; furthermore, $(a_n+1,a_i)|(a_n+1,a_n)=1$ for all $i=1,...,n$, so there is at least one prime factor inside $a_{n+1}$ which is not contained in $a_i$. Therefore, among the first $1000^{1000}$ terms, there must be with a prime factor at least $1000^{1000}$, which implies the result.

Define the sequence $(b_n)$ such that $b_1=a_1$ and $b_{n+1}=\dfrac{a_{n+1}}{a_n}$ for $n \geq 1$. Then, $a_n=b_1 b_2 \cdots b_n$ for all $n$ and so $b_{n+1}=\dfrac{a_{n+1}}{a_n}=a_n+1=b_1 \ldots b_n+1$ Thus, the prime divisors of any $b_i$ and $b_j$ with $i \neq j$ are different (indeed, if WLOG $i<j$, then $b_j=b_1 \cdots b_i \cdots b_{j-1}+1$). Hence, the sequence $L(b_n)$ is unbounded, and since $b_n \mid a_n$ for all $n$, $L(a_n)$ is unbounded, too, as desired.

Let $\omega(n)$, be the number of distinct prime factors of $n$.Example

Notice that $\gcd(a_n,a_n+1)=1$ thus since $\omega$ is multiplicative we get $\omega(a_{n+1})=\omega(a_n(a_n+1))=\omega(a_n)+\omega(a_n+1)\ge\omega(a_n)+1$ So there is at least one prime divisor that is contained in $a_{n+1}$ and not in $a_n$ and since $(a_n)_{n\ge1}$ is unbounded, there exists a large enough $k$ such that $L(a_k)>1000^{1000}$ $\blacksquare$.

Suppose FTSOC that the set of prime factors of all the $a_i$ was finite. Then the prime factors of the sequences $(a_n)$ and $(a_n+1)$ are bounded, contradicting Kobayashi.