Let $\alpha$ be irrational. There are two cases: $\alpha$ can be either algebraic or transcendental. Products of algebraic numbers are algebraic, while rational powers of transcendental numbers are transcendental. Hence, for all positive integer $n$, $\alpha^n$ is algebraic in the first case, and transcendental in the second one.
Roth proved that algebraic irrational numbers all have irrationality measure $2$. Instead, little is known about transcendental numbers. We can only say, in general, that the irrationality measure is $\geq 2$. Thus, by the previous discussion, we can conclude that:
$\alpha$ irrational algebraic $\Rightarrow$ $\mu(\alpha^n)=2$ for all integers $n \geq 1$ (in fact, this also holds for all nonzero rationals $n$, since roots of algebraic numbers are algebraic).
$\alpha$ transcendental $\Rightarrow$ $\mu(\alpha^n) \geq 2$ for all integers $n \geq 1$ (in fact, as before, this also holds for all nonzero rationals $n$).
I think that the actual value of the irrationality measure of a power of a transcendental number highly depends on the particular case. However, it is worth recalling that almost (in the sense of Lebesgue measure) all irrational numbers have irrationality measure $2$.