Example 16 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 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 $n=2$, there are $m$ ways to color $S_{1}$, and then there are $m-1$ ways to color $S_{2}$ (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, $\cdots, S_{n}$ has $m-1$ coloring methods (only ensuring that $S_{i+1}$ and $S_{i}$ are different colors, $i=1,2, \cdots, n-1$, without ensuring that $S_{n}$ and $S_{1}$ are different colors), there are a total of $m(m-1)^{n}$ coloring methods. These coloring methods can be divided into two categories:

One category is where $S_{n}$ and $S_{1}$ are different colors, and there are $a_{n}$ such methods; 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 note that $S_{n-1}$ and $S_{1}$ must be different colors, so there are $a_{n-1}$ such coloring methods. By the principle of addition, we get $a_{n}+a_{n-1}=m(m-1)^{n-1}$.
(3) Find $a_{n}$. Let $b_{n}=\frac{a_{n}}{(m-1)^{n}}$, then $b_{n}+\frac{1}{m-1} b_{n-1}=\frac{m}{m-1}$, i.e., $b_{n}-1=-\frac{1}{m-1} \cdot \left(b_{n-1}-1\right)$, which shows that $b_{n}-1$ 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 $-\frac{1}{m-1}$. Therefore, $b_{n}-1=\frac{1}{m-1}\left(-\frac{1}{m-1}\right)^{n-2}=(-1)^{n} \frac{1}{(m-1)^{n-1}}$. From this, we can get $a_{n}=(m-1)^{n} b_{n}=(m-1)^{n}+(-1)^{n}(m-1)$.