8. Find all integers \( m, n, k \) greater than 1 such that
\[
1! + 2! + \cdots + m! = n^k.
\]
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 \( m=n=3 \).