This is always true for every positive integer $n$. If $\beta^n$ were algebraic, then $\beta$ would be a root of the polynomial $X^n-\beta^n$ over the field of <algebraic number>[algebraic numbers]. By <transitivity of algebraic extensions>, $\beta$ would then be algebraic over $\mathbb Q$, contradicting its transcendence.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 4 / 7E / d / ii / Solution

 Solution (source code)