Suppose we have already proved this exponent law for when $m,n\in\mathbb{Z^+}$ as in here. Also suppose $x^{-n}=\frac{1}{x^n}$ is given as a definition.
Let $m=-\lambda$ and $n=-\gamma$, where $\lambda,\gamma\in\mathbb{Z^+}$. Then \begin{align} a^ma^n&=\frac{1}{a^\lambda}\cdot\frac{1}{a^\gamma}\tag{by definition}\\[0.5em] &= \frac{1}{a^\lambda a^\gamma}\tag{simplify}\\[0.5em] &= \frac{1}{a^{\lambda+\gamma}}\tag{by previous result}\\[0.5em] &= a^{-(\lambda+\gamma)}\tag{by definition}\\[0.5em] &= a^{-\lambda-\gamma}\tag{simplify}\\[0.5em] &= a^{m+n}.\tag{by definition} \end{align}
Does that look like a legitimate argument? Furthermore, what would the argument be if, WLOG, $m>0$ and $n<0$? If we let $n=-\gamma$, where $\gamma\in\mathbb{Z^+}$ and $\frac{a^m}{a^n}=a^{m-n}$ were already established for $m,n\in\mathbb{Z^+}$, then would \begin{align} a^ma^n&=a^ma^{-\gamma}\tag{by definition}\\[0.5em] &= \frac{a^m}{a^\gamma}\tag{by definition}\\[0.5em] &= a^{m-\gamma}\tag{by previous result}\\[0.5em] &= a^{m+n}\tag{by definition} \end{align} be a legitimate argument as well? Is there an easier way to rigorously prove exponent laws? Most of what I have seen has only related to proving these laws for when the powers are positive integers. What about when they are simply real numbers?