Example 16 Polynomial $(1-z)^{b_{1}} \cdot\left(1-z^{2}\right)^{b_{2}} \cdot\left(1-z^{3}\right)^{b_{3}} \cdots \cdots \cdot\left(1-z^{32}\right)^{b_{12}}$, where $b_{i}$ are positive integers $(i=1,2, \cdots, 32)$, and this polynomial has the following remarkable property: when expanded, and the terms of $z$ with powers higher than 32 are deleted, exactly $1-2 z$ remains. Determine $b_{32}$ (the answer can be expressed as the difference of two powers of 2).
$$
\text { Let } \begin{aligned}
f(z) & =(1-z)^{b_{1}} \cdot\left(1-z^{2}\right)^{b_{2}} \cdot\left(1-z^{3}\right)^{b_{3}} \cdots \cdots \cdot\left(1-z^{32}\right)^{b_{32}} \\
& =1-2 z+g(z),
\end{aligned}
$$

where each term of $g(z)$ is of degree higher than 32.
By comparing the coefficients of $z$ on both sides, we get $b_{1}=2$,
$$
\begin{array}{l}
f(-z)=(1+z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}}\left(1+z^{3}\right)^{b_{3}} \cdots\left(1-z^{32}\right)^{b_{32}}=1+2 z+g(-z) \text {. } \\
f(z) \cdot f(-z)=\left(1-z^{2}\right)^{b_{1}+2 b_{2}} \cdot\left(1-z^{4}\right)^{2 b_{4}} \cdot\left(1-z^{6}\right)^{b_{3}+2 b_{6}} \cdot\left(1-z^{8}\right)^{2 b_{8}} \cdot \\
\left(1-z^{10}\right)^{b_{5}+2 b_{10}} \cdot \cdots \cdot\left(1-z^{30}\right)^{b_{15}+2 b_{30}} \cdot\left(1-z^{32}\right)^{2 b_{32}} \cdot \cdots \\
=1-4 z^{2}+g_{1}(z) \text {, } \\
\end{array}
$$

where each term of $g_{1}(z)$ is of degree higher than 32 and is even.
$$
\begin{aligned}
\text { Let } & w=z^{2}, f_{1}(w)=f(z) \cdot f(-z) \text {, then } \\
f_{1}(w)= & (1-w)^{b_{1}+2 b_{2}}\left(1-w^{2}\right)^{2 b_{4}}(1-w)^{b_{3}+2 b_{6}}\left(1-w^{4}\right)^{2 b_{8}}\left(1-w^{5}\right)^{b_{5}+2 b_{10} \cdots} \\
& \left(1-w^{15}\right)^{b_{15}+2 b_{30}}\left(1-w^{16}\right)^{2 b_{32}} \cdots \\
= & 1-4 w+g_{2}(w)
\end{aligned}
$$

where each term of $g_{2}(w)$ is of degree higher than 16 with respect to $w$.
By comparing the coefficients of $w$ on both sides, we get $b_{1}+2 b_{2}=4$, hence $b_{2}=1$.
Repeating the above steps, let $r=w^{2}, f_{2}(r)=f_{1}(w) f_{1}(-w)$, we get
$$
\begin{aligned}
f_{2}(r)= & (1-r)^{b_{1}+2 b_{2}+4 b_{4}} \cdot\left(1-r^{2}\right)^{4 b_{8}} \cdot\left(1-r^{3}\right)^{b_{3}+2 b_{6}+4 b_{12}} \cdot\left(1-r^{4}\right)^{4 b_{16}} \cdot \\
& \left(1-r^{5}\right)^{b_{5}+2 b_{10}+4 b_{20}} \cdot\left(1-r^{6}\right)^{4 b_{24}} \cdot\left(1-r^{7}\right)^{b_{7}+2 b_{14}+4 b_{28}}\left(1-r^{8}\right)^{4 b_{32}} \ldots \\
= & 1-16 r+g_{3}(r),
\end{aligned}
$$

where each term of $g_{3}(r)$ is of degree higher than 8 with respect to $r$.
By comparing the coefficients of $r$ on both sides, we get $b_{1}+2 b_{2}+4 b_{4}=16$.

From $b_{1}=2, b_{2}=1$, we get $b_{4}=3$.
Continuing the above process, we can sequentially get $b_{8}=30, b_{16}=2^{12}-2^{4}=4080$. Finally, from
$$
b_{1}+2 b_{2}+4 b_{4}+8 b_{8}+16 b_{16}+32 b_{32}=2^{32} \text {, }
$$

we get $b_{32}=\frac{2^{32}-2^{16}}{32}=2^{27}-2^{11}$.