Dimension of Vector Space on Cartesian Product

Theorem
Let $\struct {K, +, \circ}$ be a division ring.

Let $n \in \N_{>0}$.

Let $\mathbf V := \struct {K^n, +, \times}_K$ be the $K$-vector space $K^n$.

Then the dimension of $\mathbf V$ is $n$.

Proof
Let the unity of $K$ be $1$, and the zero of $K$ be $0$.

Consider the vectors:

Thus $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the standard ordered basis of $\mathbf V$.

From Standard Ordered Basis is Basis, $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is a basis of $\mathbf V$.

The result follows from Linearly Independent Set is Basis iff of Same Cardinality as Dimension.