I recognise that a trivial proof exists via contradiction and complex numbers, namely: $$1=\sqrt{-1*-1}=\sqrt{-1}\sqrt{-1}=i^2=-1 $$ Which is obviously untrue therefore the rule does not hold for negative arguments of the radical function. I've been looking around for a direct proof that uses an alternate method, but I have been unable to find such a thing. Specifically, I'm asking if there is a proof that produces $\sqrt{mn} \neq \sqrt{m} \sqrt{n}$ for $m,n < 0$ without contradiction.
The reason I am interested is because if the only way to conclude the above domain for the radical identity is through contradiction, what is to say that the converse isn't true, and that the existence of complex numbers implies such 'logical flaws', and the only way to get around that is to just 'unrigorously cheat' and put a domain on the law.
In a sense I could wrap my head around that: if you think about $n^2$, because a complex number is a solution to $n^2=-1$, in a very crude way, the solutions must be of the formation of $a\times b$ where say, $b\in \Re^{-}$, but somehow maintain equality as it numbers being squared ergo $-1=1$.
Expanding on the comment of Martin R, the question is meaningless unless a convention is accepted for the definition of $\sqrt{z} ~: z \in \Bbb{C}, z \neq 0.$
A common Complex Analysis convention, which dovetails with the corresponding Real Analysis convention, is that if $z \neq 0$ and
$z = re^{i\theta} = r[\cos(\theta) + i\sin(\theta)] ~: ~-\pi < \theta \leq \pi$,
then $\sqrt{z} = \sqrt{r}e^{i\theta/2}.$
Here the principal Argument of $z = \theta$, and the principal Argument of $\sqrt{z} = \theta/2.$
Under this convention, you will always have that $-\pi/2 < \text{the principal Argument of} ~\sqrt{z} \leq \pi/2.$
Now consider $z_1 = r_1e^{i\theta_1}, z_2 = r_2e^{i\theta_2} ~:$
$0 < r_1, r_2, ~-\pi < \theta_1, \theta_2 \leq \pi.$
Then $z_1 \times z_2 = r_1r_2e^{i(\theta_1 + \theta_2)}$.
Then $\sqrt{z_1 z_2} = \sqrt{r_1 r_2} \times e^{i\alpha}$,
where $-\pi/2 < \alpha \leq \pi/2$ and $\alpha \equiv (\theta_1/2 + \theta_2/2) \pmod{\pi}.$
Contrast this with $\sqrt{z_1} \times \sqrt{z_2} = \sqrt{r_1 r_2} \times e^{i/2(\theta_1 + \theta_2)}$.
So, $\sqrt{z_1 z_2} = \sqrt{z_1} \times \sqrt{z_2} \iff (\theta_1/2 + \theta_2/2) = \alpha$.
Stating the assertion a different way, if $\theta_1, \theta_2$ are the principal Arguments of $z_1, z_2$,
then $\sqrt{z_1z_2} = \sqrt{z_1} \times \sqrt{z_2} \iff -\pi/2 < (1/2)(\theta_1 + \theta_2) \leq \pi/2.$