The answer to both questions is yes.
As a preliminary, let's prove that for any infinite-dimensional vector space $V$, that
- Lemma: $card(V) = card(k) \cdot \dim V$
Proof: Since $card(k) \leq card(V)$ and $\dim V \leq card(V)$, the inequality
$$card(k) \cdot \dim V \leq card(V)^2 = card(V)$$
is obvious. On the other hand, any element of $V$ is uniquely of the form $\sum_{j \in J} a_j e_j$ for some finite subset $J$ of (an indexing set of) a basis $B$ and all $a_j$ nonzero. So an upper bound of $card(V)$ is $card(P_{fin}(B)) \sup_{j \in P_{fin}(B)} card(k)^j$. If $B$ is infinite, then $card(P_{fin}(B)) = card(B) = \dim(V)$, and for all finite $j$ we have $card(k^j) \leq card(k)$ if $k$ is infinite, and $card(k^j) \leq \aleph_0$ if $k$ is finite, and either way we have
$$card(V) \leq \dim V \cdot \max\{card(k), \aleph_0\} \leq \dim V \cdot card(k)$$
as desired. $\Box$
The rest is now easy. Suppose $I$ is an infinite set, and suppose without loss of generality that $V_i$ is nontrivial for all $i \in I$. Put $V = \prod_{i \in I} V_i$. We have
$$\dim V \geq \dim k^I = card(k)^I \geq card(k)$$
where the equality is due to Erdos and Kaplansky. Therefore
$$\dim(V) = \dim(V)^2 \geq \dim V \cdot card(k) = card(V)$$ card(V) = \prod_i card(V_i)$$
by the lemma above.
