$10 \cdot 73$ For any positive integer $k$, let $f_{1}(k)$ be defined as the square of the sum of the digits of $k$. For $n \geqslant 2$, let $f_{n}(k)=f_{1}\left(f_{n-1}(k)\right)$. Find $f_{1988}(11)$.
(6th American Invitational Mathematics Examination, 1988)
［Solution］From the definition, we can obtain
$$
\begin{array}{l}
f_{1}(11)=(1+1)^{2}=2^{2}=4, \\
f_{2}(11)=f_{1}(4)=4^{2}=16, \\
f_{3}(11)=f_{1}\left(f_{2}(11)\right)=f_{1}(16)=(1+6)^{2}=49, \\
f_{4}(11)=f_{1}\left(f_{3}(11)\right)=f_{1}(49)=(4+9)^{2}=169, \\
f_{5}(11)=f_{1}\left(f_{4}(11)\right)=f_{1}(169)=(1+6+9)^{2}=256, \\
f_{6}(11)=f_{1}\left(f_{5}(11)\right)=f_{1}(256)=(2+5+6)^{2}=169, \\
f_{7}(11)=f_{1}\left(f_{6}(11)\right)=f_{1}(169)=(1+6+9)^{2}=256 .
\end{array}
$$

From the above, we can see that for $n \geqslant 4, \quad f_{n}(11)$ alternates between 169 and 256.
That is, for $n \geqslant 4$,
$$
f_{n}(11)=\left\{\begin{array}{ll}
169, & \text{when } n \text{ is even,} \\
256, & \text{when } n \text{ is odd.}
\end{array}\right.
$$

Therefore, $f_{1988}(11)=169$.