Example 10 (9th "Hope Cup" Invitational Training Question) The inverse function of the function $y=f(x+1)$ defined on the set of real numbers $\mathbf{R}$ is $y=f^{-1}(x+1)$, and $f(1)=3997$, then the value of $f(1998)$ is $\qquad$ .
Solve: Fill in 2000. Reason: From the given $y=f^{-1}(x+1)$, we get
$f(y)=x+1$, which means $x=f(y)+1$, thus the inverse function is $y=f(x)-1$.
From the given, we have $f(x)-1=f(x+1)$, that is $f(x+1)-f(x)=-1$.
Therefore, $f(1998)=[f(1998)-f(1997)]+[f(1997)-f(1996)]+\cdots+[f(2)-$ $f(1)]+f(1)=-1997+3997=2000$.