Let us take two natural numbers, for example $a_1=2$ and $a_2=7$. Multiply to obtain $a_3=2 \cdot 7=14$. Obtain $a_4$ as $a_2 \cdot a_3=98$. Then $a_5$ as $a_3 \cdot a_4=14 \cdot 98=1372$. Repeat this procedure an infinite number of times by defining $a_{n+2}=a_{n+1} \cdot a_n$ for $n \in \mathbb N$.

From three multiplications in this example we obtained these numbers $2,7,14,98,1372$.

If we take all possible subsets of digits of these numbers we obtain a set $\{1,2,3,4,7,8,9,12,13,14,17,32,37,72,98.132,137,172,372,1372\}$

I would like to know:

If we choose some two naturals $a_1$ and $a_2$ such that $a_1 \cdot a_2$ does not end in zero and take all possible subsets of digits of numbers in the set $\{a_{n+2}=a_{n+1} \cdot a_n\mid n\in \mathbb N\} \cup \{a_1\} \cup \{a_2\}$ do we have that set of all possible subsets of digits of numbers in the set $\{a_{n+2}=a_{n+1} \cdot a_n\mid n\in \mathbb N\} \cup \{a_1\} \cup \{a_2\}$ equals $\mathbb N$?

This seems to be way too trivial property of these sets, that is, that they all equal $\mathbb N$, so I expect some clever argument that settles this question of mine.

This is not a result of some research of some well-known research problems and topics but just a result of some thinking about some problem from recreational mathematics.