The purpose of this community wiki answer is to repeat, with a few infinitesimal changes, Todd Trimble's wonderful answer.

**Lemma.** If $V$ is a vector space which is infinite as a set, then we have $card(V)=card(k)\dim(V)$.

*Proof.* It is easy to see that, if $S$ is an infinite generating subset of a group $G$, then $S$ and $G$ are equipotent. Putting 
$$
G:=V,\qquad S:=\{\lambda b\ |\  (\lambda,b)\in k\times B\},
$$
where $B$ is a basis of $V$, we get the conclusion. 

**Theorem.** Let $I$ be an infinite set, $(V_i)_{i\in I}$ a family of nonzero vector spaces, and $V$ the product of the $V_i$. Then we have $\dim(V)=card(V)$.

*Proof.* We have 
$$
card(k)\le card\left(k^I\right)=\dim\left(k^I\right)\le\dim(V),\tag1
$$

the equality following from the [Erdős-Kaplansky Theorem](http://math.stackexchange.com/a/61537/660), and also 
$$
card(V)=card(k)\dim(V)=\dim(V),
$$ 
the first equality following from the lemma, and the second from (1).