As the title says $[1-x\_n]^{n}\to 1$ implies that $n x\_n\to 0$ as $n\to\infty$?

We know that $\left(1+\frac{x}{n} \right)^n \to e^x$ as $n\to\infty$. This implies (not sure why) that $\left(1+\frac{x}{n} +o(1/n) \right)^n \to e^x$.

So the result obtains since we must have that $x\_n=\frac{x}{n}+o(1/n)$ where $x=0$?