66. Equation in factorials. Find all solutions of the equation

$$
n!(n-1)!=m!
$$
66. We have $n!(n-1)!=[(n-1)!]^2=m$! Obviously, $1!0!=1$! and $2!1!=2!$ are solutions to this equation. In all other cases, if $n \neq m$, then there are no solutions*. Further, for $m>10$, there always exist two prime numbers $p$ and $q$ such that $>\frac{m}{2}$ and $\leqslant m$. In this case,

$$
p q \geqslant\left(\frac{m}{2}+\frac{1}{2}\right)\left(\frac{m+1}{2}+2\right)=\frac{m^2}{4}+\frac{3 m}{2}+\frac{5}{4}>m
$$

Therefore, for $m>10$ there are no solutions. For $m \leqslant 10$, apart from the two solutions mentioned above, there is only one more solution $7!6!=10!$