Prove the following:
Let $n$ be a natural number bigger or equal to $2$. Let $G_1,G_2,\ldots,G_n$ be groups. Let $(g_1,g_2,\ldots,g_n)$ be an elemement of $G_1\times G_2\times\cdots\times G_n$. The order of $(g_1,g_2,\ldots,g_n)$ is $\operatorname{lcm}(\operatorname{ord}(g_1), \operatorname{ord}(g_2),\ldots,\operatorname{ord}(g_n))$.
I understand the proof for the case $n=2$, but I can't prove the case with $n$ larger than $2$. I've tried to use proof by induction on $n$, using $n=2$ as the "base case", but I get stuck on the induction step. Any hint?
The order of $(g_1,\ldots,g_n) \in G_1 \times \cdots \times G_n$ is defined as the smallest (non-zero) integer $m \in \Bbb{N}^*$ such that $ (g_1,\ldots,g_n)^m = (g_1^m,\ldots,g_n^m) = (e_1,\ldots,e_n), $ hence $m \,|\, \mathrm{ord}(g_i) \;\forall i = 1,\ldots,n$. The smallest number dividing a set of integers corresponds precisely to the least common multiple, that is why one can conclude that $m = \mathrm{lcm}(\mathrm{ord}(g_1),\ldots,\mathrm{ord}(g_n))$.