Prove Corollary (5.6).
Corollary (5.6) states that if the integers $m_j$ are pairwise prime then
$$ \phi (m_1 × m_2 × \ldots × m_k) = \phi (m_1) × \phi (m_2) × \ldots × \phi (m_k) $$
We will prove this by induction.
Let $S(k)$ be the statement:
$$ m_j \text{ pairwise prime } \implies \phi (m_1 × m_2 × \ldots × m_k) = \phi (m_1) × \phi (m_2) × \ldots × \phi (m_k) $$
We need to prove the base case $S(2)$ and the inductive step $S(k) \implies S(k+1)$.
Base Case $S(2)$
The base case is
$$ m_1, m_2 \text{ pairwise prime } \implies \phi (m_1 × m_2) = \phi (m_1) × \phi (m_2) $$
This is Theorem (5.5), and so the base case is true.
Inductive Step $S(k) \implies S(k+1)$
We assume $S(k)$, the induction hypothesis, and aim to show $S(k+1)$.
We start assuming $m_j$ all pairwise prime,
$$ \phi ( (m_1 \times m_2 \times \ldots \times m_k) \times m_{k+1}) = \phi (m_1 \times m_2 \times \ldots \times m_k) \times \phi (m_{k+1}) $$
This is justified by $m_{k+1}$ being coprime to $(m_1 \times m_2 \times \ldots \times m_k)$.
$S(k)$ true gives us $\phi (m_1 \times m_2 \times \ldots \times m_k) = \phi (m_1) \times \phi (m_2) \times \ldots × \phi (m_k)$, and so
$$ \phi ( m_1 \times m_2 \times \ldots \times m_k \times m_{k+1}) = \phi (m_1) \times \phi (m_2) \times \ldots × \phi (m_k) \times \phi (m_{k+1}) $$
This is $S(k+1)$ and so the induction step is proven.
By induction we have shown that if $m_j$ are pairwise prime, then
$$ \phi (m_1 × m_2 × \ldots × m_k) = \phi (m_1) × \phi (m_2) × \ldots × \phi (m_k) $$