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, and also $$ card(V)=card(k)\dim(V)=\dim(V), $$ the first equality following from the lemma, and the second from (1).
EDIT 1. To stress the punchline, one can phrase things as follows.
Let $V$ be an infinite dimensional vector space.
Say that $V$ is large if $\dim(V)\ge card(k)$.
By the lemma, $V$ is large if and only if $\dim(V)=card(V)$.
Erdős-Kaplansky Theorem: $k^{\mathbb N}$ is large.
Now it's clear that if $V=\prod V_i$ with $V_i\neq0$ for infinitely many $i$'s, then $V$ is large.
EDIT 2. For the sake of completeness let us express explicitly the dimension of a product $\prod_{i\in I}V_i$ of $k$-vector spaces in terms of the $d_i:=\dim V_i$.
We can assume that $I$ is infinite and that $d_i\ge1$ for all $i$. Setting $$ \kappa:=card(k),\quad\mu:=\max(\aleph_0,\kappa),\quad\alpha:=card(\{i\in I\ |\ d_i < \mu\}), $$ we get $$ \dim\prod_{i\in I}V_i=\kappa^\alpha\prod_{d_i\ge\mu}d_i. $$