Example 6. (IMO- $22-6$ )
The function $f(x, y)$ satisfies for all non-negative integers $x, y$
$$
\begin{array}{l}
\text { (1) } f(0, y)=y+1 \text {, } \\
\text { (2) } f(x+1,0)=f(x, 1) \text {. } \\
\text { (3) } f(x+1, y+1)=f(x, f(x+1, y)) \text {. }
\end{array}
$$

Determine $f(4,1981)$.
$$
\begin{aligned}
f(1, n) & =f(0, f(1, n-1)) \\
& =f(1, n-1)+1
\end{aligned}
$$

Given $f(1,0)=f(0,1)=2$,
we get $\quad f(1, n)=n+f(1,0)=n+2$.
Also,
$$
\begin{aligned}
f(2, n) & =f(1, f(2, n-1)) \\
& =f(2, n-1)+2=2 n+f(2
\end{aligned}
$$

Since $f(2,0)=f(1,1)=3$,
we have $f(2, n)=2 n+3$.
Furthermore,
$$
\begin{aligned}
f(3, n) & =f(2, f(3, n-1)) \\
& =2 f(3, n-1)+3 \\
& =2[f(3, n-1)+3]-3
\end{aligned}
$$

Thus, $\frac{f(3, n)+3}{f(3, n-1)+3}=2$,
which implies $\boldsymbol{f}(3, \boldsymbol{n})+3=2^{n}[\boldsymbol{f}(3,0)+3]$.
Since $f(3,0)=f(2,1)=5$,
we have $f(3, n)=2^{n+3}-3$.
Finally, we calculate $f(4, n)$. H
$$
\begin{aligned}
f(4, n) & =f(3, f(4, n-1)) \\
& =2^{f(4, n-1)+3}-3
\end{aligned}
$$

Thus, $f(4, n)+3=2^{f(4, n-1)+3}$.
Let $t_{n}=f(4, n)+3, \quad p(x)=2^{x}$,

then $t_{n}=p\left(t_{n-1}\right)$.
Therefore, $t_{n}=\phi^{(n)}\left(t_{0}\right)=\varphi\left(\varphi \cdots\left(\varphi\left(t_{0}\right)\right)\right)$,

Now,
$$
\begin{aligned}
t_{0} & =f(4,0)+3 \\
& =f(3,1)+3 \\
& =2^{4}-3+3 \\
& =16
\end{aligned}
$$

So, $f(4, n)=\phi^{(n)}(16)-3$.
$$
\text { Hence, } f(4,1981)=p^{(1981)}(16)-3=2^{1984}-3 \text {. }
$$

The main exponent is 1984. Methods such as the fixed point method, similar function method, etc., can be used.