Simplify $\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} * \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$ to $\frac{\sqrt{mnc}}{a^9cmn}$ I need to simplify $$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot  \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$$
The solution provided is:  $\dfrac{\sqrt{mnc}}{a^9cmn}$.
I'm finding this challenging. I was able to make some changes but I don't know if they are on the right step or not:
First, I am able to simplify the left fractions numerator and the right fractions denominator:
$\sqrt{mn^3}=\sqrt{mn^2n^1}=n\sqrt{mn}$
$\sqrt{m^2c^4}=m\sqrt{c^2c^2} = mcc$
So my new expression looks like:
$$\frac{n\sqrt{mn}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{mcc}$$
From this point I'm really at a loss to my next steps. If I multiply them both together I get:
$$\frac{(n\sqrt{mn})(a^{-7}n^{-2})}{(a^2\sqrt{c^{-3}})(mcc)}.$$
Next, I was thinking I could multiply out the radical in the denominator but I feel like I need to simplify what I have before going forwards.
Am I on the right track? How can I simplify my fraction above in baby steps? I'm particularly confused by the negative exponents.
How can I arrive at the solution $\dfrac{\sqrt{mnc}}{a^9cmn}$?

$$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}}\cdot\frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}} = \frac{\sqrt{mn^3}a^{-7}n^{-2}}{a^2\sqrt{m^2c^{-3}c^4}}$$
From here, you use the following identity:
$$a^{-b} = \frac{1}{a^b}$$
You can simplify from here:
$$= \frac{\sqrt{mn^3}}{a^9n^2\sqrt{m^2c}} = \frac{\sqrt{n^3}}{a^9n^2\sqrt{mc}} = \color{blue}{\frac{\vert n\vert\sqrt{n}} {a^9n^2\sqrt{mc}}} = \color{green}{\frac{\sqrt{nmc}}{a^9cm\sqrt{n^2}}} = \frac{\sqrt{nmc}}{a^9cm\vert n\vert}$$
Notice the step highlighted in blue. Clearly, when dealing with real numbers, the radicand must be non-negative (and the denominator can’t be $0$), so $mc > 0$. Therefore, in the next step, we can note that $nmc > 0$, and since $mc > 0$, then $n > 0$, so the absolute value of $n$ is $n$ itself. (Mathematically, $\vert n\vert = n$.) Hence, the final result becomes
$$\frac{\sqrt{nmc}}{a^9cmn}$$
Addition: This is, of course, not to make the problem seem more confusing than it actually is. However, it is a common error to forget the absolute value sign when dealing with even indices. $\sqrt{a^2}$ is not $a$, it’s $\vert a\vert$ because $a$ itself may be negative, but the returned value is always non-negative. For example, $\sqrt{(-2)^2} = \sqrt{4} = 2 = \vert -2\vert$. (This also applies to all even indices, such as fourth roots, sixth roots, etc. Always be careful when dealing with these.)
