Let the number of color schemes for $n$ sectors be $a_n$.

When $n=2$, one sector has $m$ choices and the other has $(m-1)$ choices. Hence, $a_2=m(m-1)$.

Next let's show $$a_n + a_{n-1}=m(m-1)^{n-1}$$

for $n\ge 2$. Let these $n$ sectors be $S_1$, $S_2$, $\cdots$, $S_n$. Then, $S_1$ has $m$ different colors to choose and $S_2$ has $(m-1)$ choices. Meanwhile, $S_3$ has $(m-1)$ colors available to it which are different from $S_2$. Generally speaking, $S_{k}$ always have $(m-1)$ options to pick.

Hence, $n$ sectors have $m(m-1)^{n-1}$ such coloring schemes. However, this scheme cannot guarantee $S_n$ and $S_1$ have different colors. Among the $m(m-1)^{n-1}$ possibilities, there are $a_n$ cases when $S_1$ and $S_n$ are colored differently by definition. The rest are those situations when $S_1$ and $S_n$ have the same color. In this case, it must be true that $S_1$ and $S_{n-1}$ are colored differently. Hence, removing sector $S_n$ will construct a circle of $(n-1)$ sectors with no adjacent sections having the same color. There are $a_{n-1}$ different such $(n-1)$ sectors. In conclusion, we find $$a_n + a_{n-1}=m(m-1)^{n-1},\qquad a_2=m(m-1)$$

In order to solve $a_n$ from this recursion, let $b_n=\frac{a_n}{(m-1)^n}$. Then $$b_n+\frac{b_{n-1}}{(m-1)}=\frac{m}{m-1}\implies (b_n-1)=-\frac{1}{m-1}(b_{n-1}-1)$$

This means that $\{b_n-1\}$ is a geometric sequence whose common ratio is $-\frac{1}{m-1}$ and initial term $$b_2=\frac{a_2}{(m-1)^2} -1 =\frac{m(m-1)}{(m-1)^2}-1=\frac{1}{m-1}$$

So $$b_n = \frac{1}{m-1}\left(-\frac{1}{m-1}\right)^{n-2}={\frac{(-1)^n}{(m-1)^{n-1}}}$$

It follows that the answer is $$a_n=b_n(m-1)^n=\boxed{(m-1)^n+(-1)^n(m-1)}$$