Saudi Arabia Mathematical Competitions 2012

For $n=3$ we have $3!\cdot 5!=6!$.

Assume that $a_1!a_2!\dots a_{k-1}!=a_k!$. Since $a_k!(a_k!-1)!=(a_k!)!$, it follows that $$a_1!a_2!\dots a_{k-1}!(a_k!-1)!=a_k!(a_k!-1)!=(a_k!)!$$ and obviously $a_k!>a_k!-1>a_{k-1}!$. We are done by induction.