Consider a sequence $x_1, x_2, x_3, \cdots$ defined by
$x_1=\sqrt[3]{3}$
$x_2=(\sqrt[3]{3})^{\sqrt[3]{3}}$
and in general
$x_n=(x_{n-1})^{\sqrt[3]{3}}$ for $n>1$.
What is the smallest value of $n$ for which $x_n$ is an [integer](https://artofproblemsolving.com/wiki/index.php/Integer)?
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \  } 4 \qquad \mathrm{(D) \  } 9 \qquad \mathrm{(E) \  }27$
First, we will use induction to prove that $x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}$
We see that $x_1=\sqrt[3]{3}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^0\right)}$. This is our base case.
Now, we have $x_n=(x_{n-1})^{\sqrt[3]{3}}=\left(\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-2}\right)}\right)^{\sqrt[3]{3}}=\sqrt[3]{3}^{\left(\sqrt[3]{3}\cdot\sqrt[3]{3}^{n-2}\right)}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}$. Thus the induction is complete.
We now get rid of the cube roots by introducing fractions into the exponents. 
$x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}=\sqrt[3]{3}^{\left(3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(\frac{1}{3}\cdot3^{\left(\frac{n-1}{3}\right)}\right)}=3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)}$.
Notice that since $3$ isn't a perfect power, $x_n$ is integral if and only if the exponent, $3^{\left(\frac{n-4}{3}\right)}$, is integral. By the same logic, this is integral if and only if $\frac{n-4}{3}$ is integral. We can now clearly see that the smallest positive value of $n$ for which this is integral is $4, \boxed{\text{C}}$.