Example 1. Given $f(n)=n^{4}+n^{3}+n^{2}+n+1$, find the remainder of $f\left(2^{5}\right)$ divided by $f(2)$.
$$
\begin{aligned}
f(2) & =2^{4}+2^{3}+2^{2}+2+1 \\
& =(11111)_{2}
\end{aligned}
$$
(The subscript 2 outside the parentheses indicates a binary number, the same applies below).
$$
\begin{array}{l}
f\left(2^{5}\right)=2^{20}+2^{15}+2^{10}+2^{5}+1 \\
=(100001000010000100001)_{2} \\
=(1111100 \cdots 0)_{15 \text { ones }}+(1111100 \cdots 0)_{12} \\
+(1111100 \cdots 0 \underbrace{2}_{6 \text { zeros }}+(1111100000)_{2} \\
+(1111100)_{2}+(101)_{2} \text {, } \\
\end{array}
$$
$\therefore f\left(2^{5}\right)$ modulo $f(2)$ is
$$
(101)_{2}=2^{2}+1=5 \text {. }
$$