$8 \cdot 58$ Find the positive integers $m$, $n$, $k$ greater than 1 that satisfy $1!+2!+3!+\cdots+m!=n^{k}$.
(Canadian Mathematical Olympiad Training Problem, 1988)
［Solution］Let $f(m)=1!+2!+\cdots+m!$.
When $m>1$,
$$
3 \mid f(m) \text {. }
$$

Therefore,
$$
3 \mid n \text {. }
$$

For $m \geqslant 9, \quad m!$ is divisible by $27=3^{3}$.
But $f(8)=46233$ is not divisible by 27, so for $m \geqslant 8$, $k=2$.
Also, for $m \leqslant 6$, $f(m)$ is not divisible by 27.
$f(7)$, although divisible by 27, is not in the form of $n^{k}(k>1)$.
So it must be that $k=2$, i.e., $f(m)$ is a perfect square.
For $m>4, \quad m!$ is divisible by 5.
So for $m \geqslant 4$,
$$
f(m) \equiv f(4) \equiv 3(\bmod 5) .
$$

Thus, $f(m)$ is not a perfect square. (A perfect square modulo 5 is 0, 1, or 4).
Therefore,
$$
m \leqslant 3 .
$$

It is easy to find that there is only one set of positive integers $m, n, k$ greater than 1, which is
$$
m=3, \quad n=3, \quad k=2
$$

satisfying the given equation.