Suppose we want to prove that some number $x$ is transcendental. Can we do so by proving that $$x^n\ne\frac{a}{b}$$ where $a, b, n \in \Bbb{Z}$ where $n \ne 0$? The reasoning behind the statement is as follows:
Consider a polynomial equation: $0 = x^n + f(x)$ where $f(x)$ is a polynomial function where $f(x) \in \Bbb Q$ if $x \in \Bbb Q$. We can rearrange the polynomial as $-f(x)=x^n$. The LHS is rational and therefore the RHS is rational for rational values of x. We can solve the equation by taking the $nth$ root of both sides: $$\sqrt[n]{-f(x)}=x$$
The LHS and RHS now may be irrational, but remain algebraic. Therefore, if $x$ is transcendental, then $x^n\ne\frac{a}{b}$