Call the numbers $a_1, a_2, \cdots, a_{2022}$ As $a_ia_{i+1}$ is perfect cube for all odd $i$, we know that $\prod_{i=1}^{1011} a_{2i-1}a_{2i}=\prod_{i=1}^{2022}a_i$ is perfect cube. As $a_ia_{i+1}$ is perfect cube for all even $i$, we know that $\prod_{i=1}^{1011} a_{2i}a_{2i+1}=\prod_{i=2}^{2021}a_i$ is perfect cube. So $a_1a_{2022}$, being the quotient of the two expressions, is a perfect cube.