Prove that $V^n$ and $\mathcal{L}(\mathbf{F}^n,V)$ are isomorphic vector spaces 
For $n$ positive integer, define $V^n$ by $V^n=\underbrace{V\times...\times V}_{n \ times}$. Prove that $V^n$ and $\mathcal{L}(\mathbf{F}^n,V)$ are isomorphic vector spaces. I would like to know if my proof holds and to have a feedback, please. ($\mathbf{F}$ denotes a field here)

Let $(v_1,...,v_n)$ be a basis of $V$. So, each element in $V$ can be expressed as $\lambda_1 v_1+...+\lambda_n v_n$ for $\lambda_1,...,\lambda_n \in \mathbf{F}$.
Let $\xi:\mathbf{F}^n\to V$, $\xi(\lambda_1,...,\lambda_n)=\lambda_1 v_1+...+\lambda_n v_n$ and define $\psi: V^n\to \mathcal{L}(\mathbf{F}^n,V)$ as $\psi (\lambda_1 v_1+...+\lambda_n v_n,...,\lambda_1 v_1+...+\lambda_n v_n)=\xi(\lambda_1,...,\lambda_n)$.
Clearly $\psi$ is a linear application (it is easy to check). We show now that $\psi$ is injective.
$\psi(\lambda_1 v_1+...+\lambda_n v_n,...,\lambda_1 v_1+...+\lambda_n v_n)=\xi(\lambda_1,...,\lambda_n)=\lambda_1v_1+..+\lambda_nv_n=0 \iff \lambda_1=...=\lambda_n=0$ because $(v_1,...,v_n)$ is linearly independent in $V$. So, $\lambda_1 v_1+...+\lambda_n v_n,...,\lambda_1 v_1+...+\lambda_n v_n=0$ and we conclude that $\psi$ is injective.
Moreover, the dimension of $V^n$  is equal to a dimension of $\mathcal{L}(\mathbf{F}^n,V)$. Thus, by fundamental theorem we conclude that $\psi$  is surjective. Therefore, $\psi$ is an isomorphism

In fact the result is true whatever the dimension of $V$ is.
Consider
$$\begin{array}{l|rcl}
\Phi : & V^n & \longrightarrow & \mathcal L(F^n,V)\\
    & (v_1,\dots,v_n) & \longmapsto & (\lambda_1, \dots, \lambda_n) \mapsto \lambda_1v_1+ \dots + \lambda_n v_n\end{array}$$
$\Phi$ is linear, injective as its kernel is the set consisting of the zero vector and surjective.
