(i) Prove that if $a \mid c$ and $b \mid c$ and $\gcd (a, b) = 1$ then $(a \times b) \mid c$.
(ii) Prove that if $a_1 \mid c, a_2 \mid c, \ldots , a_n \mid c$ and $\gcd (a_j, a_i) = 1$ where $i \neq j$ then
$$(a_1 \times a_2 \times \ldots \times a_n) \mid c$$
(i) We're given $a \mid c$ and $b \mid c$, which means for some integers $j,k$
$$\begin{align} c &= ja \\ \\ c &= kb \end{align}$$
We're also given $\gcd(a,b) = 1$. By Bezout's Indentity, there are integers $x,y$ such that
$$ ax + by = 1 $$
Multiplying by $c$,
$$ ax(kb) + by(aj) = c $$
Factorising,
$$ab(xk + yj) = c$$
And so we conclude $(a \times b) \mid c$
(ii) We'll prove this by induction on $n$.
Let's take the statement $P(n)$ to mean:
If $a_1 \mid c, a_2 \mid c, \ldots , a_n \mid c$ and $\gcd (a_j, a_i) = 1$ where $i \neq j$ then $(a_1 \times a_2 \times \ldots \times a_n) \mid c$.
We need to prove a base case and an induction step.
Base Case
The base case $P(2)$ is
If $a_1 \mid c, a_2 \mid c$ and $\gcd (a_1, a_2) = 1$ where $1 \neq 2$ then $(a_1 \times a_2) \mid c$.
We proved this base case in exercise (i) above.
Induction Step
We need to show $P(n) \implies P(n+1)$.
We assume $P(n)$ as the induction hypothesis.
The statement $P(n+1)$ means:
If $a_1 \mid c, a_2 \mid c, \ldots , a_n \mid c, a_{n+1}\mid c$ and $\gcd (a_j, a_i) = 1$ where $i \neq j$ then $(a_1 \times a_2 \times \ldots \times a_n \times a_{n+1}) \mid c$.
The assumption for $P(n+1)$ that $a_1 \mid c, a_2 \mid c, \ldots , a_n \mid c, a_{n+1}\mid c$ where all $a_j, a_{i \ne j}$ are coprime, includes the assumption for $P(n)$ that $a_1 \mid c, a_2 \mid c, \ldots , a_n \mid c$ where all $a_j, a_{i \ne j}$ are coprime. That means we have $(a_1 \times a_2 \times \ldots \times a_n) \mid c$.
That is, there exists some integer $p$ such that
$$(a_1 \times a_2 \times \ldots \times a_n) p= c$$
The additional assumption of $P(n+1)$ is that $a_{n+1} \mid c$, so there exists some integer $q$ such that
$$ a_{n+1}q = c$$
Since $a_{n+1}$ is coprime to all $a_1, a_2, \ldots, a_n$, by Bezout's identity we have, for some integers $r,s$
$$ (a_1 \times a_2 \times \ldots \times a_n) r + a_{n+1} s = 1 $$
Multiplying by $c$
$$ (a_1 \times a_2 \times \ldots \times a_n) ra_{n+1}q + a_{n+1} s(a_1 \times a_2 \times \ldots \times a_n)p = c $$
Factorising
$$ (a_1 \times a_2 \times \ldots \times a_n \times a_{n+1}) (rq + sp) = c $$
So we have $(a_1 \times a_2 \times \ldots \times a_n \times a_{n+1}) \mid c$.
We have shown $P(n) \implies P(n+1)$.
By showing the base case and the induction step, we have shown by induction that if $a_1 \mid c, a_2 \mid c$ and $\gcd (a_1, a_2) = 1$ where $1 \neq 2$ then $(a_1 \times a_2) \mid c$.