Prove that this particular sequence contains an infinite number of sixes
Solution 1:
For any block of consecutive elements of the sequence, the products of consecutive pairs form another block that will appear somewhere further on. The following blocks form a cycle in this manner: $$8, 8, 8 \to 6, 4, 6, 4 \to 2, 4, 2, 4, 2, 4 \to 8, 8, 8, 8, 8$$
Since $8, 8, 8$ appears in the sequence starting at position 72, there will be an infinite number of $6$s in the sequence (as well as $2$s, $4$s, and $8$s).