The first two numbers of the $i$-th row are $i$ and $i+d$. Denote the greatest common divisor of this row by $a$; since $a$ divides both $i$ and $i+d$, it must also divide their difference $d$. Thus all the greatest common divisors divide $d$, meaning their least common multiple is at most $d$. But as the final row is $d, 2d, \dots, n$ with the greatest common divisor $d$, the least common multiple of the greatest common divisors is exactly $d$.