For all $n \in \mathbb { N } ^ { * }$, we denote by $d _ { n }$ the LCM of the natural integers between 1 and $n$, in other words: $d _ { n } = \operatorname { LCM } ( 1,2 , \ldots , n )$. For all prime number $p$, we denote by $k _ { p }$ the largest natural integer such that $p ^ { k _ { p } } \leqslant n$.
Show that $d _ { n } = \prod _ { \substack { p \leqslant n \\ p \text { prime } } } p ^ { k _ { p } }$.