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Here is a self-contained version of Todd Trimble's wonderful answer.

Let $K$ be a field. "Vector space" shall mean "$K$-vector space", "linear" shall mean "$K$-linear", $\dim$ shall mean $\dim_K$, $\operatorname{Hom}$ shall mean $\operatorname{Hom}_K$, and $|X|$ shall denote the cardinal of $X$ for any set $X$.

Let $V$ be the product of a family of nonzero vector spaces $(V_i)_{i\in I}$: $$ V=\prod_{i\in I}V_i. $$

As we have $$ \dim V=\sum_{i\in I}\dim V_i $$ if $I$ is finite, we can (and will) assume from now on that $I$ is infinite.

Main Theorem. We have, in the above notation, $\dim V=|V|$. In words: the dimension of the product of an infinite family of nonzero vector spaces is equal to its cardinal.

As a corollary, let us express explicitly the dimension of the product $V$ of the $V_i$ in terms of the $d_i:=\dim V_i$. Setting $$ \mu:=\max(\aleph_0,|K|),\quad\alpha:=|\{i\in I\ |\ d_i < \mu\}|, $$ we get $$ \dim\prod_{i\in I}V_i=|K|^\alpha\prod_{d_i\ge\mu}d_i. $$

Let us prove the Main Theorem.

Lemma. If $V$ is a vector space which is infinite as a set, then we have $$ |V|=|K|\cdot\dim V. $$

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. QED

Let $V$ be an infinite dimensional vector space.

Say that $V$ is large if $\dim V\ge |K|$.

By the lemma, $V$ is large if and only if $\dim V=|V|$.

Erdős-Kaplansky Theorem. The vector space $K^{\mathbb N}$ is large.

It is clear that the Erdős-Kaplansky Theorem implies the Main Theorem. So we are left with proving the Erdős-Kaplansky Theorem.

Proof of the Erdős-Kaplansky Theorem. Let $B$ be a $K$-basis of $K^{\mathbb N}$, and suppose by contradiction $|B|<|K|$. Let $K_0$ be the prime subfield of $K$, and put $$ K_1:=K_0(\{b_j\ |\ b\in B,\ j\in\mathbb N\}). $$ As $|K_1|<|K|$ and $K$ is infinite, there is an $x$ in $K^{\mathbb N}$ whose coordinates are $K_1$-linearly independent. There are $c_1,\dots,c_n$ in $B$ such that $x$ is a $K$-linear combination of the $c_j$. Since $c_{ij}$ is in $K_1$ for all $i,j$, there is a nonzero $\lambda$ in $K_1^{n+1}$ such that $$ \sum_{j=0}^n\lambda_j\,c_{ij}=0 $$ for $1\le i\le n$, and we have $$ \sum_{j=0}^n\lambda_j\,x_j=0, $$ in contradiction with the choice of $x$. QED

Here is a self-contained version of Todd Trimble's wonderful answer.

Let $K$ be a field. "Vector space" shall mean "$K$-vector space", "linear" shall mean "$K$-linear", $\dim$ shall mean $\dim_K$, $\operatorname{Hom}$ shall mean $\operatorname{Hom}_K$, and $|X|$ shall denote the cardinal of $X$ for any set $X$.

Let $V$ be the product of a family of nonzero vector spaces $(V_i)_{i\in I}$: $$ V=\prod_{i\in I}V_i. $$

As we have $$ \dim V=\sum_{i\in I}\dim V_i $$ if $I$ is finite, we can (and will) assume from now on that $I$ is infinite.

Main Theorem. We have, in the above notation, $\dim V=|V|$. In words: the dimension of the product of an infinite family of nonzero vector spaces is equal to its cardinal.

As a corollary, let us express explicitly the dimension of the product $V$ of the $V_i$ in terms of the $d_i:=\dim V_i$. Setting $$ \mu:=\max(\aleph_0,|K|),\quad\alpha:=|\{i\in I\ |\ d_i < \mu\}|, $$ we get $$ \dim\prod_{i\in I}V_i=|K|^\alpha\prod_{d_i\ge\mu}d_i. $$

Let us prove the Main Theorem.

Lemma. If $V$ is a vector space which is infinite as a set, then we have $$ |V|=|K|\cdot\dim V. $$

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. QED

Let $V$ be an infinite dimensional vector space.

Say that $V$ is large if $\dim V\ge \max (|K|, \aleph_0)$.

By the lemma, $V$ is large if and only if $\dim V=|V|$.

Erdős-Kaplansky Theorem. The vector space $K^{\mathbb N}$ is large.

It is clear that the Erdős-Kaplansky Theorem implies the Main Theorem. So we are left with proving the Erdős-Kaplansky Theorem.

Proof of the Erdős-Kaplansky Theorem. Let $B$ be a $K$-basis of $K^{\mathbb N}$, and suppose by contradiction $|B|<|K|$. Let $K_0$ be the prime subfield of $K$, and put $$ K_1:=K_0(\{b_j\ |\ b\in B,\ j\in\mathbb N\}). $$ As $|K_1|<|K|$ and $K$ is infinite, there is an $x$ in $K^{\mathbb N}$ whose coordinates are $K_1$-linearly independent. There are $c_1,\dots,c_n$ in $B$ such that $x$ is a $K$-linear combination of the $c_j$. Since $c_{ij}$ is in $K_1$ for all $i,j$, there is a nonzero $\lambda$ in $K_1^{n+1}$ such that $$ \sum_{j=0}^n\lambda_j\,c_{ij}=0 $$ for $1\le i\le n$, and we have $$ \sum_{j=0}^n\lambda_j\,x_j=0, $$ in contradiction with the choice of $x$. QED

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 $$ \dim\prod_{i\in I}V_i=\kappa^\alpha\prod_{d_i\ge\mu}d_i. $$

Here is a self-contained version of Todd Trimble's wonderful answer.

Let $K$ be a field. "Vector space" shall mean "$K$-vector space", "linear" shall mean "$K$-linear", $\dim$ shall mean $\dim_K$, $\operatorname{Hom}$ shall mean $\operatorname{Hom}_K$, and $|X|$ shall denote the cardinal of $X$ for any set $X$.

Let $V$ be the product of a family of nonzero vector spaces $(V_i)_{i\in I}$: $$ V=\prod_{i\in I}V_i. $$

As we have $$ \dim V=\sum_{i\in I}\dim V_i $$ if $I$ is finite, we can (and will) assume from now on that $I$ is infinite.

Main Theorem. We have, in the above notation, $\dim V=|V|$. In words: the dimension of the product of an infinite family of nonzero vector spaces is equal to its cardinal.

As a corollary, let us express explicitly the dimension of the product $V$ of the $V_i$ in terms of the $d_i:=\dim V_i$. Setting $$ \mu:=\max(\aleph_0,|K|),\quad\alpha:=|\{i\in I\ |\ d_i < \mu\}|, $$ we get $$ \dim\prod_{i\in I}V_i=|K|^\alpha\prod_{d_i\ge\mu}d_i. $$

Let us prove the Main Theorem.

Lemma. If $V$ is a vector space which is infinite as a set, then we have $$ |V|=|K|\cdot\dim V. $$

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. QED

Let $V$ be an infinite dimensional vector space.

Say that $V$ is large if $\dim V\ge |K|$.

By the lemma, $V$ is large if and only if $\dim V=|V|$.

Erdős-Kaplansky Theorem. The vector space $K^{\mathbb N}$ is large.

It is clear that the Erdős-Kaplansky Theorem implies the Main Theorem. So we are left with proving the Erdős-Kaplansky Theorem.

Proof of the Erdős-Kaplansky Theorem. Let $B$ be a $K$-basis of $K^{\mathbb N}$, and suppose by contradiction $|B|<|K|$. Let $K_0$ be the prime subfield of $K$, and put $$ K_1:=K_0(\{b_j\ |\ b\in B,\ j\in\mathbb N\}). $$ As $|K_1|<|K|$ and $K$ is infinite, there is an $x$ in $K^{\mathbb N}$ whose coordinates are $K_1$-linearly independent. There are $c_1,\dots,c_n$ in $B$ such that $x$ is a $K$-linear combination of the $c_j$. Since $c_{ij}$ is in $K_1$ for all $i,j$, there is a nonzero $\lambda$ in $K_1^{n+1}$ such that $$ \sum_{j=0}^n\lambda_j\,c_{ij}=0 $$ for $1\le i\le n$, and we have $$ \sum_{j=0}^n\lambda_j\,x_j=0, $$ in contradiction with the choice of $x$. QED

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. 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.

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 $$ \dim\prod_{i\in I}V_i=\kappa^\alpha\prod_{d_i\ge\mu}d_i. $$