18.4.11 * Find the integer solution of the equation $\left(1+\frac{1}{m}\right)^{m+1}=\left(1+\frac{1}{1988}\right)^{1988}$.
The original equation is transformed into $\frac{1989^{1988}}{1988^{1988}}=\frac{(m+1)^{m+1}}{m^{m+1}}$.
If $m$ is a positive integer, then since both sides of (1) are irreducible fractions, it must be that
$$
m^{m+1}=1988^{1988}.
$$

However, when $m \geqslant 1988$, $m^{m+1}>1988^{1988}$. When $0<m<1988$, $m^{m+1}<1988^{1988}$. Therefore, there is no solution.
Clearly, $m \neq 0$. When $m=-1$, the left side of the original equation is 0, so $m=-1$ is not a solution.
When $m<-1$, let $n=-(m+1)$. (1) becomes
$$
\frac{1989^{1988}}{1988^{1988}}=\left(\frac{-(n+1)}{-n}\right)^{n}=\frac{(n+1)^{n}}{n^{n}},
$$

Thus, $1988^{1988}=n^{n}, n=1988 . m=-1989$.
The problem has exactly one solution $m=-1989$.