I want to prove that $\sqrt[7]{129}$ is irrational using the following theorem:
Let $n,k$ be natural numbers. Then, $\sqrt[n]{k}$ is rational iff $k$ is the $n\text{-th}$ power of a (natural) number.
Attempt: Let $\sqrt[7]{129}\in\mathbb{Q}.$ So, there are positive integers $m,n$ with $GCF(m,n)=1$ such that $\sqrt[7]{129}=\frac{m}{n}.$ If $n>1,$ there is a prime $p$ such that $p\mid n,$ thus, $129n^7=m^7.$ So, $p\mid m^7=m\cdot m\cdots m\Leftrightarrow p\mid m.$ That doesn't holds because $GCF(m,n)=1.$ So $n=1$ and $129=m^7.$ Now, $m$ has to be an odd number, let it be $m=2k+1$ for some $k\in\mathbb{N}.$ $\text{Or,}$ $$129=(2k+1)^7.$$ (I really have No idea what step follows next. Any help would be appreciated.)
$2^7=128$, $3^7>3^5=243>129$. Hence $m$ cannot be an integer.