Suppose $V_1$ and $V_2$ are $k$-vector spaces with bases $(e_i)_{i \in I}$ and $(f_j)_{j \in J}$, respectively. I've seen the claim that the collection of elements of the form $e_i \otimes f_j$ (with $\left(i,j\right) \in I \times J$) forms a basis for $V_1 \otimes V_2$. But I seem to get stuck with the proof.
My question: What's the easiest way to see that the above set is indeed linearly independent?
Construct a set $\{\phi_{i,j}\}$ of linear forms on the tensor product which is a dual basis to the family $\{e_i \otimes f_j\}$ (in the sense that for each pair $\left(i, j\right) \in I \times J$, the map $\phi_{i,j}$ sends $e_i \otimes f_j$ to $1$ while sending all $e_{i'} \otimes f_{j'}$ with $\left(i', j'\right) \neq \left(i, j\right)$ to $0$). That will immediately imply linear independence.
If $\{\psi_i\}$ is a dual basis to your basis $\{e_i\}$ of $V_1$ and $\{\rho_j\}$ is a dual basis to your basis $\{f_j\}$ of $V_2$, then you can consider the map $\phi_{i,j} = \psi_i\otimes \rho_j:V_1\otimes V_2\to k\otimes k\cong k$ for each pair $\left(i, j\right) \in I \times J$. This map $\phi_{i,j}$ sends $e_i \otimes f_j$ to $1$ while sending all $e_{i'} \otimes f_{j'}$ with $\left(i', j'\right) \neq \left(i, j\right)$ to $0$.