$13 \cdot 62$ Consider the sequence $x_{1}, x_{2}, x_{3}, \cdots$, where $x_{1}=\sqrt[3]{3}, x_{2}=$ $(\sqrt[3]{3})^{\sqrt[3]{3}}$, and generally, $x_{n}=\left(x_{n-1}\right)^{\sqrt[3]{3}}$. For $n>1$, what is the smallest $n$ such that $x_{n}$ is an integer?
(A) 2 .
(B) 3 .
(C) 4 .
(D) 9 .
(E) 27 .
(36th American High School Mathematics Examination, 1985)
[Solution] Let $x_{1}=\sqrt[3]{3}=3^{\frac{1}{3}}, x_{2}=(\sqrt[3]{3})^{\sqrt[3]{3}}=\left(3^{\frac{1}{3}}\right)^{3^{\frac{1}{3}}}=3^{\frac{1}{3} \cdot 3^{\frac{1}{3}}}$, $x_{3}=x_{2}^{\sqrt[3]{3}}=3^{\frac{1}{3} \cdot 3^{\frac{1}{3}} \cdot 3^{\frac{1}{3}}}, \cdots$
In general, $x_{n}=3^{\frac{1}{3} \cdot \underbrace{\frac{1}{3}}_{n} \cdot 3^{\frac{1}{3}} \ldots 3^{\frac{1}{3}}}=3^{3^{-1+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\cdots+\frac{1}{3}}}$.
Thus, if $x_{n}$ is an integer for the smallest $n$ which is 4.
That is,
$$
x_{4}=3^{3^{-1+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}}=3^{3^{0}}=3^{1}=3 .
$$

Therefore, the answer is $(C)$.