6. The factorial of a positive integer $m$ is the product of all positive integers up to $m$, denoted as $m!$ (for example, $2!=1 \times 2=2$, $(3!)!-(1 \times 2 \times 3)!=6!=1 \times 2 \times 3 \times 4 \times 5 \times 6=720$). If $((n!)!)!$ is a factor of $(2021!)!$, what is the maximum value of $n$? $\qquad$ .
【Answer】6
【Analysis】For natural numbers $a, b$, $a$ ! is a factor of $b$ ! if and only if $a \leq b$. Since (( $n!)$ !)! is a factor of (2021!)!, we have $(n!)!\leq 2021!$, which means $n!\leq 2021$. $6!-1 \times 2 \times 3 \times 4 \times 5 \times 6=720,7!-7 \times 720>2021$, so the maximum value of $n$ is 6.