4-168 Find the integer solutions of the equation
$$
\left(1+\frac{1}{m}\right)^{m+1}=\left(1+\frac{1}{1988}\right)^{1988} \text {. }
$$
[Solution] The original equation can be transformed into
$$
\frac{(m+1)^{m+1}}{m^{m+1}}=\frac{1989^{1988}}{1988^{1988}}
$$

If $m>0$, then both sides of (1) are irreducible fractions, so it must be that
$$
m^{m+1}=1988^{1988}
$$

Notice that when $m \geqslant 1988$, $m^{m+1}>1988^{1988}$;
$$
01988$ when, $n^{n}>1988^{1988}$;
$$
0<n<1988 \text { when, } n^{n}<1988^{1988} \text {; }
$$

Therefore, (4) has only one solution $n=1988$. After calculation, we get
$$
m=-1989 \text {. }
$$

Upon verification, the integer solution to the original equation is $m=-1989$.

Note: There seems to be a typo in the original text where "01988" appears, which is likely a formatting or input error. The correct interpretation should be "0 < n < 1988".