How to show that an $n$-dimensional vector space over field $\mathbb{F}_p$ has $p^n$ elements. Suppose $V$ is an $n$-dimensional vector space over the finite field $\mathbb{F}_p$ for some prime $p$. How do I show that $V$ has $p^n$ elements? I was thinking that considering $n$ basis elements one can show that each element of the basis spans its own subspace which doesn't have any other basis elements, but how do I show that there are exactly $p^n$ elements overall?

Let $(e_1,\dots,e_n)$ be a base of $V$ as a $\mathbb{F}_p$-vector space. By definition of a basis, the application
$$
\begin{array}{rcl}(\mathbb{F}_p)^n & \longrightarrow & V\\(x_1,\dots,x_n) & \longmapsto & x_1e_1 + \dots + x_n e_n\end{array}
$$
yields an isomorphism $V \simeq (\mathbb{F}_p)^n$. In particular, $\text{card}V = \text{card}(\mathbb{F}_p)^n =  p^n$.
