Nth Power of a Natural Number Introduces no New Prime Factors

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Theorem

The nth power of any natural number has the same prime factors as the natural number.

Proof

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 $x^n$

$\blacksquare$