12.75 Find the integer solution $m$ of the equation $\left(1+\frac{1}{m}\right)^{m+1}=\left(1+\frac{1}{1988}\right)^{1988}$.
(22nd All-Soviet Union Mathematical Olympiad, 1988)
[Solution] (1) If $m>0$, then the given equation becomes
$$
\frac{(m+1)^{m+1}}{m^{m+1}}=\frac{1989^{1988}}{1988^{1988}}.
$$

Since $(m, m+1)=1$ and $(1988,1989)=1$, both sides of this equation are in their simplest form, so their denominators are equal.
$$
m^{m+1}=1988^{1988}
$$

If $m \geqslant 1988$, then $m^{m+1}>1988^{1988}$,
If $0<m<1988$, then $m^{m+1}<1988^{1988}$.

Therefore, $m=1988$.

(2) If $m<0$, let $m=-(n+1)$.
The given equation becomes
$$
\begin{aligned}
\left(1-\frac{1}{n+1}\right)^{-n} & =\left(1+\frac{1}{1988}\right)^{1988}, \\
\frac{(n+1)^{n}}{n^{n}} & =\frac{1989^{1988}}{1988^{1988}}
\end{aligned}
$$

Thus, $n=1988$, and $m=-1989$.