2, ** 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 the requirement that adjacent sectors must have different colors. How many different coloring methods are there?
Let the number of coloring methods be $a_{n}(m)$.
(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 $a_{2}(m)=m(m-1)$.
(2) Find the recurrence relation, since $S_{1}$ has $m$ coloring methods, $S_{2}$ has $m-1$ coloring methods, $\cdots, S_{n-1}$ has $m-1$ coloring methods, and $S_{n}$ still has $m-1$ coloring methods (not ensuring that $S_{n}$ and $S_{1}$ are different colors), there are a total of $m(m-1)^{n-1}$ coloring methods. However, 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 $a_{n}(m)$, and the other category is where $S_{n}$ and $S_{1}$ are the same color. In this case, merging $S_{n}$ and $S_{1}$ into one sector and noting that $S_{n-1}$ and $S_{1}$ must be different colors, the number of coloring methods is $a_{n-1}(m)$. By the principle of addition, we get
$$
a_{n}(m)+a_{n-1}(m)=m(m-1)^{n-1} \quad (n \geqslant 2).
$$
(3) Find $a_{n}(m)$, let $b_{n}(m)=\frac{a_{n}}{(m-1)^{n}}$, then from (1) we get $b_{n}(m)+\frac{1}{m-1} b_{n-1}(m)=\frac{m}{m-1}$, i.e., $b_{n}(m)-1=-\frac{1}{m-1}\left(b_{n-1}(m)-1\right)$, so
$$
\begin{aligned}
b_{n}(m)-1 & =\left(b_{2}(m)-1\right) \cdot\left(-\frac{1}{m-1}\right)^{n-2}=\left[\frac{a_{2}}{(m-1)^{2}}-1\right]\left(-\frac{1}{m-1}\right)^{n-2} \\
& =\left[\frac{m(m-1)}{(m-1)^{2}}-1\right]\left(-\frac{1}{m-1}\right)^{n-2}=(-1)^{n} \frac{1}{(m-1)^{n-1}},
\end{aligned}
$$

Therefore, $a_{n}(m)=(m-1)^{n} b_{n}(m)=(m-1)^{n}+(-1)^{n}(m-1)$. Thus, there are $(m-1)^{n}+(-1)^{n}(m-1)$ different coloring methods.