Let $G_1, G_2, ... , G_n$ be finite cyclic groups. I want to show that $G_1 \times G_2 \times ... \times G_n$ is cyclic if, and only if, $\gcd(|G_i||G_j|) = 1$ whenever $i \ne j$. I do know that if $G$ and $H$ are finite cyclic groups, then $G \times H$ is cyclic if, and only if, $\gcd(|G||H|)=1$. I'm assuming that some form of induction would work when dealing with $n$ finite cyclic groups, but I have absolutely no idea how to apply it.
Any help would be much appreciated.
You're exactly right, you need to use induction. I'm going to write $m_i:=|G_i|$; so in one direction, if $\gcd(m_i,m_j)=1$ for $i\neq j$, then by the induction hypothesis you have that $G_1\times\cdots\times G_{n-1}$ is cyclic. Then using the result for $n=2$ (which you say you already know), you just need to show that $\gcd(m_1\cdots m_{n-1},m_n)=1$ to deduce that $(G_1\times\cdots\times G_{n-1})\times G_n$ is cyclic as well.
On the other hand, if $G_1\times\cdots\times G_n$ is cyclic, recall a quotient of a cyclic group is cyclic, so $G_1\times\cdots\times G_{n-1}$ and $G_n$ are cyclic groups, and by the induction hypothesis on the former you see that $\gcd(m_i,m_j)=1$ whenever $1\le i,j\le n-1$ and $i\neq j$. But then again you can use the $n=2$ case to also deduce that $\gcd(m_1\cdots m_{n-1},m_n)=1$, and putting these two facts together find that $\gcd(m_i,m_j)=1$ whenever $1\le i,j\le n$ and $i\neq j$.