Solve the system in positive numbers:

$$
\begin{cases}x^{y} & =z \\ y^{z} & =x \\ z^{x} & =y\end{cases}
$$
First, note that if one of the unknowns is equal to one, then the others are also equal to one.

Indeed, let $x=1$. Then $z=1^{\mathrm{y}}=1, y=z^{\mathrm{x}}=1^{1}=1$. Suppose there exists another solution besides $(1,1,1)$. Let's first consider $x>1$. Then $z=x^{\mathrm{y}}>1, y=z^{\mathrm{x}}>1$. Therefore, $z=x^{\mathrm{y}}>x^{1}=x, y=z^{\mathrm{x}}>z^{1}=z, x=y^{\mathrm{z}}>y^{1}=y$, which is impossible. Thus, this case is impossible. The case $x<1$ is impossible for the same reason by replacing all the ">" signs with "<" signs.

## Answer

$x=y=z=1$.