6. Let $0<a<1$, if $x_{1}=a, x_{2}=a^{x_{1}}, x_{3}=a^{x_{2}}, \cdots, x_{n}=a^{x_{n-1}}, \cdots$, then the sequence $\left\{x_{n}\right\}$
A. is increasing
B. is decreasing
C. odd terms are increasing, even terms are decreasing
D. even terms are increasing, odd terms are decreasing
Solve:
$0<a<1 \Rightarrow y=a^{x}$ is monotonically decreasing $\Rightarrow a^{1}<a^{a}<a^{0} \Rightarrow x_{1}<x_{2}<1$
$$
\begin{array}{l}
\Rightarrow a^{1}<a^{x_{2}}<a^{x_{1}} \Rightarrow x_{1}<x_{3}<x_{2} \Rightarrow a^{x_{2}}<a^{x_{3}}<a^{x_{1}} \Rightarrow x_{3}<x_{4}<x_{2} \\
\Rightarrow a^{x_{2}}<a^{x_{4}}<a^{x_{3}} \Rightarrow x_{3}<x_{5}<x_{4} \Rightarrow a^{x_{4}}<a^{x_{5}}<a^{x_{3}} \Rightarrow x_{5}<x_{6}<x_{4} \\
\Rightarrow a^{x_{4}}<a^{x_{6}}<a^{x_{5}} \Rightarrow x_{5}<x_{7}<x_{6} \Rightarrow \cdots,
\end{array}
$$

Thus $0<x_{1}<x_{3}<x_{5}<\cdots<x_{2 n-1}<\cdots<x_{2 n}<\cdots<x_{6}<x_{4}<x_{2}<1$.
Therefore, the odd terms are increasing, and the even terms are decreasing, so the answer is $C$.