6. The number of pairs of positive integer solutions $(n, m)$ for the equation $1!+2!+3!+\cdots+n!=m$ ! is $\qquad$ pairs.
Obviously, $m=n=1$ is a solution; when $n \geqslant 2$, since $n!<m!<(n+1)!=(n+1) \cdot n!$, there is no solution, so the only pair $(n, m)$ that satisfies the condition is one pair.