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

This can be false. For any nonzero transcendental number $\beta$, the number $\gamma =1/\beta$ is also transcendental, since otherwise its reciprocal would be algebraic. Yet $\beta \gamma =1$ is algebraic.

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 numbers. By transitivity of algebraic extensions, $\beta$ would then be algebraic over $\mathbb{Q}$, contradicting its transcendence.