Example 1 Divide a circle into $n(\geqslant 2)$ sectors $S_{1}, S_{2}, \cdots, S_{n}$. Now, color these sectors using $m(\geqslant 2)$ colors, with each sector being colored with exactly one color and requiring that adjacent sectors have different colors. How many different coloring methods are there?
Let the number of coloring methods be $a_{n}$.
(1) Find the initial value. When $n=2$, coloring $S_{1}$ has $m$ methods, and then coloring $S_{2}$ has only $m-1$ methods (since $S_{1}$ and $S_{2}$ must be different colors), so $a_{2}=m(m-1)$.
(2) Find the recurrence relation. Since $S_{1}$ has $m$ coloring methods, $S_{2}$ has $m-1$ coloring methods, $S_{3}$ has $m-1$ coloring methods, $\cdots, S_{n-1}$ has $m-1$ coloring methods, and $S_{n}$ still has $m-1$ coloring methods (not guaranteeing that $S_{n}$ and $S_{1}$ are different colors). This results in a total of $m(m-1)^{n-1}$ coloring methods, but these $m(m-1)^{n-1}$ coloring methods can be divided into two categories:

One category is where $S_{n}$ and $S_{1}$ are different colors, in which case the number of coloring methods is $a_{n}$; the other category is where $S_{n}$ and $S_{1}$ are the same color, in which case $S_{n}$ and $S_{1}$ can be merged into one sector, and noting that $S_{n-1}$ and $S_{1}$ must be different colors, the number of coloring methods in this case is $a_{n-1}$. By the principle of addition, we get
$$
a_{n}+a_{n-1}=m(m-1)^{n-1}(n \geqslant 2) .
$$
(3) Find $a_{n}$. Let $b_{n}=\frac{a_{n}}{(m-1)^{n}}$, from $b_{n}+\frac{1}{m-1} b_{n-1}=\frac{m}{m-1}$, i.e., $b_{n}-1=-\frac{1}{m-1}\left(b_{n-1}-\right.$ 1), i.e., $b_{n}-1$ is the $(n-1)$-th term of a geometric sequence with the first term $b_{2}-1=\frac{a_{2}}{(m-1)^{2}}-1=\frac{1}{m-1}$ and the common ratio $-\frac{1}{m-1}$,
so $b_{n}-1=\left(\frac{1}{m-1}\right)\left(-\frac{1}{m-1}\right)^{n-2}=(-1)^{n} \frac{1}{(m-1)^{n-1}}$,

i.e., $\frac{a_{n}}{(m-1)^{n}}=1+(-1)^{n} \frac{1}{(m-1)^{n-1}}$.
Therefore, there are $a_{n}=(m-1)^{n}+(-1)^{n}(m-1)$ coloring methods.