Example 9 The function $f(x, y)$ satisfies for all non-negative integers: (1) $f(0, y)=y+1$; (2) $f(x+1,0)=$ $f(x, 1)$; (3) $f(x+1, y+1)=f(x, f(x+1, y))$. Determine $f(4,1981)$.
From condition (3), it indicates that when $f(x, y)$ is known, a difference equation for $f(x+1, y)$ in terms of $y$ can be given. Condition (2) provides the initial value. Since condition (1) gives the expression for $f(0, y)$, we have for $f(1, y)$: $f(1, y+1)=f(0, f(1, y))=f(1, y)+1, f(1,0)=f(0,1)=2$. Solving this difference equation yields $f(1, y)=y+2$. Thus, we can derive the difference equation and initial value for $f(2, y)$:
$f(2, y+1)=f(1, f(2, y))=f(2, y)+2, f(2,0)=f(1,1)=3$. Solving this difference equation gives $f(2, y)=2 y+3$. Then, we derive the difference equation and initial value for $f(3, y)$:
$f(3, y+1)=f(2, f(3, y))=2 f(3, y)+3, f(3,0)=f(2,1)=5$.
Solving this difference equation yields $f(3, y)=2^{y+3}-3$. Thus, we derive the difference equation and initial value for $f(4, y)$: $f(4, y+1)=f(3, f(4, y))=2^{f(4, y)+3}-3, f(4,0)=f(3,1)=2^{4}-3$. Solving this difference equation gives $f(4, y)=2^{2^{y+3}-3}$, so $f(4,1981)=2^{2^{1984}-3}$.