Prove: (arithmetic mean $=$ geometric mean) $\Rightarrow$ all variables are equal for $n>2$

This answer presumes $\text{AM}\ge\text{GM}$ is true. In other words, we consider $\text{AM}\ge\text{GM}$ as a black box, which we uses without proof. If we already know some proof of this inequality, however, the result would immediately follow from it.
Now suppose for some $x_1\ne x_2$ we have $\text{AM}(x_1,x_2,\cdots,x_n)=\text{GM}(x_1,x_2,\cdots,x_n)$. Let $x_1'=x_2'=(x_1+x_2)/2$, we see $$\text{AM}(x_1',x_2',\cdots,x_n)=\text{AM}(x_1,x_2,\cdots,x_n)$$ Meanwhile, we have $x_1'x_2'-x_1x_2=(x_1-x_2)^2/4>0$, hence $$\text{GM}(x_1',x_2',\cdots,x_n)>\text{GM}(x_1,x_2,\cdots,x_n)$$ Combining these results, we find $$\text{AM}(x_1',x_2',\cdots,x_n)<\text{GM}(x_1',x_2',\cdots,x_n)$$ Contradicting $\text{AM}-\text{GM}$ inequality. So $x_1=x_2$ if the equality holds. For the same reason, all $x_i$ is equal.

Here is a hint for the proof of $n=4$

$(x_1x_2x_3x_4)^{1/4} = \sqrt{\sqrt{x_1x_2}\sqrt{x_3x_4}}$

Can you extend that to $n=8$ and so on?

Then, if you have it for $n=4$, you can show it for $n=3$ by letting $x_4$ be the average of the other three.