In this book: Lang, Serge. "Basic Mathematics" (p. 71), appears this theorem and proof.
Theorem 1. Let $a$, $b$ be positive real numbers. Then, $$(ab)^{1/n}=a^{1/n}b^{1/n}$$
Proof. Let $r=a^{1/n}$ and $s=b^{1/n}$. This means that $r^n=a$ and $s^n=b$. Therefore $$ \begin{align} (rs)^n =r^ns^n = ab \end{align} $$
- How does the author came up with the equality $(rs)^n =r^ns^n$ ? That exponent property wasn't introduced in the book at this point.
If $n$ is a positive integer and $r$ and $s$ are any positive real numbers, then we find that $$ \begin{align} (rs)^n &= \underbrace{(rs) \times \cdots \times (rs)}_{n \mbox{ times} } \\ &= \left( \underbrace{ r \times \cdots \times r}_{n \mbox{ times} } \right) \times \left( \underbrace{ s \times \cdots \times s}_{n \mbox{ times} } \right) \\ &= r^n \times s^n. \end{align} $$ Here we have used the commutative property of multiplication.
Alternatively, we can prove $$ (rs)^n = r^n s^n $$ for all positive real numbers $r$ and $s$ and for all natural numbers $n$ using induction on $n$.
Hope this helps.