When n and m are both positive integers, to show that $a^n a^m = a^{n+m}$ will be pretty straightforward:
$$a^n a^m = (\underbrace{a\cdot a\cdots a}_{n\text{ copies}})(\underbrace{a\cdot a\cdots a}_{m\text{ copies}}) = (\underbrace{a\cdot a\cdots a}_{n+m\text{ copies}})= a^{n+m}$$
Now suppose that, for example, $n \in \mathbb R$ and $0 < n < 1$ and $m \in \mathbb R$
How do you show that $a^na^m = a^{n+m}$ in that scenario too?