Nth Power of a Natural Number Introduces no New Prime Factors
Let $x$ be a natural number.
Let $n$ be a natural number.
Suppose that $x$ has a set of unique prime factors, such that $x = p_1p_2 \cdots p_k$.
When $x$ is raised to the $n$th power, each prime factor is raised to the $n$th power.
|\(\displaystyle x^n\)||\(=\)||\(\displaystyle (p_1p_2 \cdots p_k)^n\)|
|\(\, \displaystyle \implies \, \)||\(\displaystyle x^n\)||\(=\)||\(\displaystyle p_1^np_2^n \cdots p_k^n\)|
Thus, raising $x$ to the $n$th power introduces no new prime factors since each prime factor in $x$ appears $n$ times in $x^n$