smc

Firstly, we will show by induction that $$x_n={\left(\sqrt[3]{3}\right)}^{\left({\left(\sqrt[3]{3}\right)}^{n-1}\right)}.$$ For the base case, we indeed have $$\begin{array}{rl}{x}_1& =\sqrt[3]{3}\\ & ={\left(\sqrt[3]{3}\right)}^1\\ & ={\left(\sqrt[3]{3}\right)}^{\left({\left(\sqrt[3]{3}\right)}^0\right)},\end{array}$$ and for the inductive step, if our claim is true for $x_n$, then $$\begin{array}{rl}{x}_{n+1}& ={\left(x_n\right)}^{\sqrt[3]{3}}\\ & ={\left({\left(\sqrt[3]{3}\right)}^{\left({\left(\sqrt[3]{3}\right)}^{n-1}\right)}\right)}^{\sqrt[3]{3}}\\ & ={\left(\sqrt[3]{3}\right)}^{\left({\left(\sqrt[3]{3}\right)}^{n-1}\cdot \sqrt[3]{3}\right)}\\ & ={\left(\sqrt[3]{3}\right)}^{\left({\left(\sqrt[3]{3}\right)}^n\right)},\end{array}$$ which completes the proof. We now rewrite our formula for $x_n$ as follows: $$\begin{array}{rl}{x}_n& ={\left(\sqrt[3]{3}\right)}^{\left({\left(\sqrt[3]{3}\right)}^{n-1}\right)}\\ & ={\left(\sqrt[3]{3}\right)}^{\left(3^{\frac{n-1}{3}}\right)}\\ & =3^{\left(\frac{1}{3}\cdot 3^{\frac{n-1}{3}}\right)}\\ & =3^{\left(3^{\left(\frac{n-1}{3}-1\right)}\right)}\\ & =3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)},\end{array}$$ and as $3$ is not a perfect power, we deduce that $x_n$ is an integer if and only if the exponent, $3^{\left(\frac{n-4}{3}\right)}$, is itself an integer. By precisely the same argument, this reduces to $\frac{n-4}{3}$ being an integer, so the smallest possible (positive) value of $n$ is $\boxed{4}$.