How to prove $\lim_{n \to \infty} \sqrt{n}(\sqrt[n]{n} - 1) = 0$?

The OP's attempt can be pushed to get a complete proof. $$ n = (1+x_n)^n \geq 1 + nx_n + \frac{n(n-1)}{2} x_n^2 + \frac{n(n-1)(n-2)}{6} x_n^3 > \frac{n(n-1)(n-2) x_n^3}{6} > \frac{n^3 x_n^3}{8}, $$ provided $n$ is "large enough" ¹. Therefore, (again, for large enough $n$,) $x_n < 2 n^{-2/3}$, and hence $\sqrt{n} x_n < 2n^{-1/6}$. Thus $\sqrt{n} x_n$ approaches $0$ by the sandwich (squeeze) theorem.

---

¹In fact, you should be able to show that for all $n \geq 12$, we have $$ \frac{n(n-1)(n-2)}{6} > \frac{n^3}{8} \iff \left( 1-\frac1n \right) \left( 1- \frac2n \right) \geq \frac34. $$

Use the fact that, when $n\to\infty$, $$\sqrt[n]{n}-1=\exp\left(\frac{\log n}n\right)-1\sim\frac{\log n}n$$

An elementary proof using $\text{AM} \ge \text{GM}$:

We have that, for sufficiently large $n$,

$$ \frac{1 + 1 + \dots + 1 + n^{1/3} + n^{1/3} + n^{1/3}}{n} \ge n^{1/n}$$

using $\text{AM} \ge \text{GM}$ on $n-3$ copies of $1$ and three copies of $n^{1/3}$.

i.e we get the estimate

$$ 1 - \frac{3}{n} + \frac{3}{n^{2/3}} \ge n^{1/n}$$

This proof can be generalized to show that

$$n^{(k-1)/k} (n^{1/n} - 1) \to 0$$

for any positive integer $k$.

A variant of: Proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$