Prove Proposition (5.9).
Proposition (5.9) states that if $n = p_1^{k_1} \times p_2^{k_2} \times \ldots \times p_r^{k_r}$, then
$$ \phi(n) = n \bigl( 1 - \frac{1}{p_1} \bigr) \bigl( 1 - \frac{1}{p_2} \bigr) \ldots \bigl( 1 - \frac{1}{p_r} \bigr) $$
We'll start from Proposition (5.9) which states that for $n>1$, if its prime decomposition is $n = p_1^{k_1} \times p_2^{k_2} \times \ldots \times p_r^{k_r}$, where $p_i$ are distinct primes, then
$$ \phi(n) = \bigl( p_1^{k_1} - p_1^{k_1-1} \bigr) \times \bigl( p_2^{k_2} - p_2^{k_2-1} \bigr) \times \ldots \times \bigl( p_r^{k_r} - p_r^{k_r -1} \bigr) $$
And so
$$ \begin{align} \phi(n) & = \bigl( p_1^{k_1} - p_1^{k_1-1} \bigr) \times \bigl( p_2^{k_2} - p_2^{k_2-1} \bigr) \times \ldots \times \bigl( p_r^{k_r} - p_r^{k_r -1} \bigr) \\ \\ & = p_1^{k_1}\bigl( 1 - p_1^{-1} \bigr) \times p_2^{k_2} \bigl( 1 - p_2^{-1} \bigr) \times \ldots \times p_r^{k_r}\bigl( 1 - p_r^{-1} \bigr) \\ \\ & = \underbrace{\bigl[ p_1^{k_1} \times p_2^{k_2} \times \ldots \times p_r^{k_r} \bigr ]}_{n} \; \bigl( 1 - \frac{1}{p_1} \bigr) \bigl( 1 - \frac{1}{p_2} \bigr) \ldots \bigl( 1 - \frac{1}{p_r} \bigr) \\ \\ & = n \bigl( 1 - \frac{1}{p_1} \bigr) \bigl( 1 - \frac{1}{p_2} \bigr) \ldots \bigl( 1 - \frac{1}{p_r} \bigr) \end{align} $$