Show that $\phi (2^n) = \frac{1}{2} (2^n)$ where $n$ is a natural number.
What does $\phi (2^n) = \frac{1}{2} (2^n)$ signify?
Using Proposition (5.4) we have
$$ \phi(2^n) = 2^{n-1}(2-1) = 2^{n-1} = \frac{1}{2}(2^n)$$
This is the quantity of odd numbers (not multiples of 2) from 1 to $2^n$.