How to prove $(a \cdot b)^n = a^n \cdot b^n$?

Question

Assuming $a, b \in \{\mathbb{R^-,R^+}\}$ and $n \in \mathbb{R}$. We often see the statement: $$(a \cdot b)^n = a^n \cdot b^n$$

Why we get the statement? How to prove this?

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There should pay attention, the $n \in \mathbb{R}$, not $\mathbb{N}$.

Answer 1

For any $a>0$, is defined $a^x=e^{x\log a}$ for all $x\in \mathbb{R}$. So $$(ab)^x=e^{x\log (ab)}=e^{x(\log a +\log b)}=e^{x\log a}\cdot e^{x\log b}=a^xb^x$$ The assumption $a, b> 0$ is clearly necessary for this to be true.