How does this division work? $\frac{\;\frac{6^6}{1}\;}{2^{-3}}\cdot2^{-10}$ I came across this division and can't wrap my head around how it is solved:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}$$
They subtract the exponent of $2^{-10}$ from the denominator's exponent $2^{-3}$:
$$2^{-3-(-10)}$$
Which gives us:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^7}$$
If anyone knows what is actually being done here I would appreciate it!

First off, the way in which the problem is typeset is almost intended to cause confusion.  Adding some parentheses and not using \dfrac everywhere should make it easier to understand the expression:
$$\frac{\left(\frac{6^6}{1}\right)}{2^{-3}} \cdot 2^{-10}. $$
Division is the inverse operation of multiplication and so, rather than thinking of the problem as a division problem, it might be easier to think of it as the multiplication of several fractions.  Specifically,
$$\frac{\left(\frac{6^6}{1}\right)}{2^{-3}} \cdot 2^{-10}
= 6^6 \cdot \frac{1}{2^{-3}} \cdot 2^{-10}. $$
One of the fundamental properties of exponents is that (under appropriate hypotheses) $a^{-n} = \frac{1}{a^n}$, and so we can further simplify this expression to
$$ 6^6 \cdot \frac{1}{2^{-3}} \cdot 2^{-10}
= 6^6 \cdot 2^{3} \cdot \frac{1}{2^{10}}
= 6^6 \cdot \frac{2^{3}}{2^{10}}. $$
Finally, another property of exponents is that
$$ \frac{a^m}{a^n} = a^{m-n}. $$
Applying this to the last result gives
$$ 6^6 \cdot \frac{2^{3}}{2^{10}}
= 6^6 \cdot 2^{3-10}
= 6^6 \cdot 2^{-7}
\overset{\text{or}}{=} \frac{6^6}{2^7}. $$
Personally, I would also probably use the fact that $(ab)^n = a^n b^n$ to rewrite $6^6 = 2^6\cdot 3^6$, and conclude by writing
$$ 6^6 \cdot 2^{-7} = 2^6\cdot 3^6 \cdot 2^{-7}
= 2^{6-7} \cdot 3^6
= 2^{-1} \cdot 3^6
\overset{\text{or}}{=} \frac{3^6}{2}, $$
but perhaps this is not part of the intended exercise.
