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!

\begin{align}
\frac{\;\;\dfrac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}
&=\frac{6^6}{2^{-3}}\cdot 2^{-10}\qquad&\text{(simplify the denominator)}\\
&=6^6\cdot 2^3\cdot 2^{-10}\qquad &(\frac{1}{2^{-3}}=2^3)\\
&=6^6\cdot 2^{3+(-10)}\qquad &(2^a2^b=2^{a+b})\\
&=6^6\cdot 2^{-7}\qquad &\text{(simplify)}\\
&=\frac{6^6}{2^7}\qquad &(2^{-7}=\frac{1}{2^7})
\end{align}
