Simplify this expression to a common fraction:  $\frac{1}{\frac{1}{(\frac{1}{2})^{1}}+\frac{1}{(\frac{1}{2})^{2}}+\frac{1}{(\frac{1}{2})^{3}}}$
I recall that raising a fraction to a power means raising both the numerator and the denominator to that power, so I can rewrite each of the smaller fractions as powers of 2: $\frac{1}{\frac{1}{2}^{1}}=2^{1}=2$, $\frac{1}{\frac{1}{2}^{2}}=2^{2}=4$, and $\frac{1}{\frac{1}{2}^{3}}=2^{3}=8$. This makes the denominator of the big fraction $2+4+8=14$. So the big fraction simplifies to $\frac{1}{14}$, which is already in lowest terms. 