In Wikipedia it is written:
The Cartesian product of an infinite number of sets each containing at least two elements is either empty or infinite; if the axiom of choice holds, then it is infinite.
How can one prove that it is either empty or infinite?
My idea is that you assume it is finite but non-empty, and then consider its cartesian product with a set of at least two elements and come up with a contradiction, but I'm not sure exactly how.
Note that the definition of infinite is that for every natural number there is a subset that bijects to it.
Suppose $f\in\prod_{i\in I} A_i$, where $I$ is infinite and each $A_i$ has at least two elements. We want to show that $\prod_{i\in I} A_i$ is infinite.
To do this, fix some injection $h:\{1, 2, ..., n\}\rightarrow I$. Since each $A_i$ has at least two elements, we can find a tuple $(a_1, a_2, ..., a_n)$ with $a_k\in A_{h(k)}$ and $f(h(k))\not=a_k$. (Note that the existence of such a tuple does not require choice, since we're just making finitely many choices here. And this is the only choice-flavored step in the proof.)
Now for $k\in \{1, 2, ..., n\}$, let $f_k$ be given by
$f_k(h(k))=a_k$, and
if $i\not=h(k)$, then $f_k(i)=f(i)$.
That is, to build $f_k$ we start with $f$ and change the $h(k)$th coordinate to $a_k$.
Clearly the $f_k$ are distinct, so the map $k\mapsto f_k$ is an injection from $\{1, 2, ..., n\}$ into the product $\prod_{i\in I}A_i$. Since this works for all $n$, the latter is infinite.