jmc

We use the prime factorizations of $6!$ and $(4!{)}^2$ to find their $\text{lcm}$ (as we would with most pairs of integers): $$\begin{array}{rcrcr}6!& =& 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1& =& 2^4\cdot 3^2\cdot 5^1\\ (4!{)}^2& =& (4\cdot 3\cdot 2\cdot 1{)}^2& =& 2^6\cdot 3^2\\ \text{lcm}[6!,(4!{)}^2]& =& 2^6\cdot 3^2\cdot 5^1& =& \boxed{2880}\end{array}$$