nth-root of irrational will always give irrational numbers?

Obviously, if $p$ is rational, then $p^2$ must also be rational.
$p\in \mathbb{Q}\Rightarrow p^2\in \mathbb{Q}.$
Take the contraposition, we see that if $x$ is irrational, then $\sqrt{x}$ must also be irrational.
$p^2\notin \mathbb{Q}\Rightarrow p\notin \mathbb{Q}.$
By negative power I assume you mean ($1/n$)-th power (it is obvious that $(\sqrt{2}{)}^{-2}=\frac{1}{2}\in \mathbb{Q}$. It is true by the statement above — just replace ${}^2$ by ${}^{\mathrm{n}}$.