If $a$ is a transcendental number, then is $a^n$ also a transcendental number? [closed]

If $a$ is a transcendental number (i.e., a number s.t. there does not exist a polynomial $P(x)$ s.t. $P(a) = 0$), is $a^n$ also transcendental?

It would seem to me that it should be, but I can't figure out why. How would I prove this?

• Is $n$ supposed to be an integer? Nov 4, 2015 at 12:24
• You should specify if $n$ is supposed to be integer, rational, or something else. For example this if false if we can take $n$ real, e.g. if $a = e$ and $n = \log(2)$, then $a^n = 2$ even though $a$ is transcendental.
– A.P.
Nov 4, 2015 at 15:08

If $p(a^n)=0$ for some polynomial $p(x)$, let $q(x)=p(x^n)$ and observe that $$q(a)=p(a^n)=0.$$ Hence, if $a^n$ is algebraic, so is $a$.
If $a^n$ isn't transcendental then $a=(b)^{1/n}$ for some nontranscendental number $b\in R$ thus $f(b)=\sum\limits_{i=0}^{k}c_i b^i=0$ for some polynomial equation where $c_i\in \mathbb{Q}$. So $g(a)=f(a^n)=\sum\limits_{i=0}^{k}c_i a^{ni}=0$. So thus $a$ isn't trancendental.