B. Divide $2,3, \cdots, n(n \geqslant 2)$ into two groups arbitrarily. If it is always possible to find numbers $a, b, c$ (which can be the same) in one of the groups such that $a^{b}=c$, find the minimum value of $n$.
B. When $n=2^{16}-1$, divide $2,3, \cdots, n$ into the following two arrays:
$$
\begin{array}{l}
\left\{2,3,2^{8}, 2^{8}+1, \cdots, 2^{16}-1\right\}, \\
\left\{4,5, \cdots, 2^{8}-1\right\} .
\end{array}
$$

In the first array, by
$$
3^{3}2^{16}-1 \text {, }
$$

we know that there do not exist numbers $a, b, c$ such that $a^{b}=c$.
In the second array, by $4^{4}>2^{8}-1$, we know that there do not exist numbers $a, b, c$ such that $a^{b}=c$.
Therefore, $n \geqslant 2^{16}$.
Next, we prove that when $n=2^{16}$, the conditions of the problem are satisfied.
Assume 2 is in the first array. If $2^{2}=4$ is also in the first array, the conclusion is already established. Hence, assume $2^{2}=4$ is in the second array.

Similarly, assume $4^{4}=2^{8}$ is in the first array, and $\left(2^{8}\right)^{2}=2^{16}$ is in the second array.
At this point, consider the number 8.
If 8 is in the first array, take $a=2, b=8, c=2^{8}$, then $a^{b}=c$;

If 8 is in the second array, take $a=4, b=8, c=2^{16}$, then $a^{b}=c$.
In summary, $n=2^{16}$ satisfies the conditions of the problem.
Therefore, the minimum value of $n$ is $2^{16}$.
(Liu Jinying provided)