$9 \cdot 1$ If $a=2^{55}, b=3^{44}, c=5^{33}, d=6^{22}$, then the order from smallest to largest of $a, b, c$, $d$ is
(A) $a<b<c<d$.
(B) $a<b<d<e$.
(C) $b<a<c<d$.
(D) $a<d<b<c$.
(2nd China Tianjin "Chinese Youth Cup" Mathematics Invitational Competition, 1988)
［Solution］Given $a=\left(2^{5}\right)^{11}=32^{11}, \quad b=\left(3^{4}\right)^{11}=81^{11}$,
$$
c=\left(5^{3}\right)^{11}=125^{11}, \quad d=\left(6^{2}\right)^{11}=36^{11},
$$

Since $32^{11}<36^{11}<81^{11}<125^{11}$, we have $a<d<b<c$. Therefore, the answer is $(D)$.