$9 \cdot 8$ Given three numbers $x, y=x^{x}, \quad z=x^{x^{x}}$, where $0.9<x<1.0$, the increasing order of these numbers is
(A) $x<z<y$.
(B) $x<y<z$.
(C) $y<x<z$.
(D) $y<z<x$.
(E) $z<x<y$.
(19th American High School Mathematics Examination, 1968)
[Solution] Given $0<x<1$, hence $\frac{x}{y}=\frac{x}{x^{x}}=x^{1-x}<1$, and $\frac{z}{y}=\frac{x^{y}}{x^{x}}=x^{y-x}<1$, so $\frac{x}{z}=\frac{x}{x^{y}}=x^{1-y}<1$, which implies $x<z$. Therefore, $x<z<y$.
Thus, the answer is $(A)$.