Problem 307

8. Solution: First, we prove that when $m \geqslant 8$, it must be that $k=2$. It is known that $1!+2!+\cdots+8!$ has a factor of $3^{2}$, but not a factor of $3^{3}$. Since $9!$, $10!$, etc., all contain powers of 3 greater than 3, when $m \geqslant 8$, $1!+2!+\cdots+m!$ has a factor of $3^{2}$, but not a factor of $3^{3}$. Therefore, $k=2$. Thus, the problem is reduced to Example 7 (the difference being that in this problem, the letters take values greater than 1), from which we can deduce that $m=n=3$.