* Find the positive integer solutions of the equation: $n!+1=(-1)^{2}$.

The equation simplifies to $n!+1=1$, since $(-1)^2=1$. Therefore, $n! = 0$. The only value of $n$ that satisfies this equation is $n=0$, but since we are looking for positive integer solutions, there are no solutions.
Parse: It is easy to verify, $m \geqslant 3$. Rewrite the equation as $n!=m!(m!-2) \cdots$ (1). Clearly, $n>m$. From (1), we get $n(n \cdots 1) \cdots(m+1)=m!\cdots 2 \cdots$ (2). Since $3 \mid m!$, it follows that $3 \nmid m!-2$, hence the left side of (2) cannot have more than two factors (otherwise, the product of three consecutive natural numbers is divisible by 3). Therefore, $n-m=1$ or $n-m=2$. In the first case, we have $m+1=m!-2$, which means $m!=m+3 \leqslant 2 m$, so $m=3, n=4$. In the second case, we have
$$
m!=(m+1)(m+2)+2 \leqslant 4 m(m-1) .
$$

Thus, $m=3$ or $m=4$, with no other solutions. Upon verification, only $m=3, n=4$ is a solution to the equation.