7. For a given positive integer $k$, let $f_{1}(k)$ denote the square of the sum of the digits of $k$, and set $f_{n+1}(k)=f_{1}\left(f_{n}\right.$ $(k))(n \geqslant 1)$. Find the value of $f_{2005}\left(2^{2006}\right)$.
7. Solution: Note that $10^{m} \equiv 1(\bmod 9), \therefore$ the positive integer $\mathbf{N}=a_{n} \times 10^{n}+\cdots+a_{1} \times 10+a_{0} \equiv a_{n}+\cdots+a_{1}+$ $a_{0}(\bmod 9)$. Let $f_{1}\left(2^{2006}\right)=a_{1}^{2}$, then $a_{1} \equiv 2^{2006} \equiv$ $2^{3 \times 668} \times 4 \equiv 4(\bmod 9)$, Let $f_{2}\left(2^{2006}\right)=f_{1}\left(a_{1}^{2}\right)=$ $a_{2}^{2}$, then $a_{2} \equiv a_{1}^{2} \equiv 7(\bmod 9)$, Let $f_{3}\left(2^{2006}\right)=f_{1}\left(a_{2}^{2}\right)$ $=a_{3}^{2}$, then $a_{3} \equiv a_{2}^{2} \equiv 4(\bmod 9)$,
$$
\begin{array}{c}
\because \lg 2^{2006}=2006 \times \lg 2<700, \therefore f_{1}\left(2^{2006}\right) \\
<(9 \times 700)^{2}<4 \times 10^{7}, \\
f_{2}\left(2^{2006}\right)<(3+9 \times 7)^{2}<4500, f_{3}\left(2^{2006}\right)< \\
(3+3 \times 9)^{2}=30^{2}, \because a_{3}<30, \therefore a_{3}=4,13,22 . \\
f_{3}\left(2^{2006}\right) \in\{16,169,242\}, f_{3}\left(2^{2006}\right)=16
\end{array}
$$

When, $f_{4}\left(2^{2006}\right)=7^{2}=49, f_{5}\left(2^{2006}\right)=13^{2}=169$,
$$
\begin{array}{l}
f_{6}\left(2^{2006}\right)=16^{2}=256, f_{7}\left(2^{2006}\right)=13^{2}=169, \\
\cdots, f_{2 n}\left(2^{2006}\right)=16^{2}=256, \\
\quad f_{2 n+1}\left(2^{2006}\right)=13^{2}=169, \therefore f_{2005}\left(2^{2006}\right)=
\end{array}
$$