$f_{n+1}(k)=f_{1}\left(f_{n}(k)\right)$. Find the value of $f_{1991}\left(2^{1990}\right)$.
Let $a_{n}=f_{n}\left(2^{1990}\right)$. Since $f_{1}(k) \equiv k^{2}(\bmod 9)$, we have $a_{n}=a_{n-1}^{2}(\bmod 9)$, $n>1$.
$$
\begin{array}{c}
a_{1}=f_{1}\left(2^{1990}\right) \equiv\left(2^{1990}\right)^{2} \equiv 4(\bmod 9), a_{2}=a_{1}^{2} \equiv 7(\bmod 9), \\
a_{3} \equiv a_{2}^{2} \equiv 4(\bmod 9), a_{4} \equiv a_{3}^{2} \equiv 7(\bmod 9), \cdots,
\end{array}
$$

That is, $a_{2 m-1} \equiv 4, a_{2 m} \equiv 7(\bmod 9), m \geqslant 1$.
Since $2^{1990}<10^{600}$, we have $a_{1} \leqslant(9 \times 600)^{2}<10^{8}, a_{2}<(9 \times 8)^{2}<10^{4}, a_{3}<$ $(9 \times 4)^{2}=1296, a_{4} \leqslant(9 \times 3)^{2}=729$. From $6+9+9<25$ and $a_{4} \equiv 7(\bmod 9)$, we know that the sum of the digits of $a_{4}$ is 7 or 16. Thus, $a_{5}=49$ or 256. Therefore, $a_{6}=169, a_{7}=256, a_{8}=169, a_{9}=$ $256, \cdots$. Hence, $a_{1991}=f_{1991}\left(2^{1990}\right)=256$.