Let $V_1,\dots,V_k$ and $W$ be vector spaces. A map $F:V_1\times\cdots\times V_k\to W$ is multilinear if $$F\left(v_1,\dots,av_i+a'v_i',\dots,v_k\right)=aF\left(v_1,\dots,v_i,\dots,v_k\right)+a'F\left(v_1,\dots,v_i',\dots,v_k\right).$$ A multilinear function of two variables is bilinear. Let $V$ be a finite-dimensional real vector space and let $k$ be a natural number. A covariant $k$-tensor on $V$ is a multilinear function $$T:\underbrace{V\times\cdots\times V}_\text{k copies}\to\mathbb R.$$ The number $k$ is called the rank of $T$. The set of all covariant $k$-tensors on $V$, denoted $T^k\left(V\right)$, is a vector space under the operations of pointwise addition and scalar multiplication. Let $V$ be a finite-dimensional real vector space, let $S\in T^k\left(V\right)$, let $T\in T^l\left(V\right)$, and define a map $$S\otimes T:\underbrace{V\times\cdots\times V}_\text{k+l copies}\to\mathbb R$$ by $$S\otimes T\left(X_1,\dots,X_{k+l}\right)=S\left(X_1,\dots,X_k\right)T\left(X_{k+1},\dots,X_{k+l}\right).$$ Then $S\otimes T$ is a covariant $\left(k+l\right)$-tensor called the tensor product of $S$ and $T$.

Proposition. Let $V$ be a real vector space of dimension $n$, let $\left(E_i\right)$ be any basis for $V$, and let $\left(\varepsilon^i\right)$ be the dual basis. The set of all $k$-tensors of the form $\varepsilon^{i_1}\otimes\cdots\otimes\varepsilon^{i_k}$ for $1\leqslant i_1,\dots,i_k\leqslant n$ is a basis for $T^k\left(V\right)$, which therefore has dimension $n^k$.

Proof. Let $\mathcal B:=\left\{\varepsilon^{i_1}\otimes\cdots\otimes\varepsilon^{i_k}:1\leqslant i_1,\dots,i_k\leqslant n\right\}$, let $T\in T^k\left(V\right)$, and, for any $k$-tuple $\left(i_1,\dots,i_k\right)$ of integers $1\leqslant i_j\leqslant n$, let $$T_{i_1\dots i_k}:=T\left(E_{i_1},\dots,E_{i_k}\right).$$ Then \begin{align}T_{i_1\dots i_k}\varepsilon^{i_1}\otimes\varepsilon^{i_k}\left(E_{j_1},\dots,E_{j_k}\right)&=T_{i_1\dots i_k}\varepsilon^{i_1}\left(E_{j_1}\right)\cdots\varepsilon^{i_k}\left(E_{j_k}\right)\\&=T_{i_1\dots i_k}\delta_{j_1}^{i_1}\cdots\delta_{j_k}^{i_k}\\&=T_{j_1\dots j_k}\\&=T\left(E_{j_1},\dots,E_{j_k}\right).\end{align} Suppose $$T_{i_1\dots i_k}\varepsilon^{i_1}\otimes\varepsilon^{i_k}=0.$$ Applying this to any $\left(E_{j_1},\dots,E_{j_k}\right)$ yields that $T_{j_1\dots j_k}=0$. $\blacksquare$

Note: Einstein's summation convention is used above.

Here is a more abstract (and more insane) definition of tensor product:

Let $S$ be a set. The free vector space on $S$, denoted $\mathbb R\langle S\rangle$, is the set of all finite formal linear combinations of elements of $S$ with real coefficients. More precisely, a finite formal linear combination is a $\mathcal F:S\to\mathbb R$ such that $\mathcal F=0$ for all but finitely-many elements of $S$. Under pointwise addition and scalar multiplication, $\mathbb R\langle S\rangle$ is a real vector space. Identifying each $x\in S$ with the function that takes the value $1$ on $x$ and $0$ elsewhere, any $\mathcal F\in\mathbb R\langle S\rangle$ can be written uniquely in the form $F=\sum_{i=1}^ma_ix_i$, where $x_1,\dots,x_m$ are the elements of $S$ for which $F\left(x_i\right)=a_i$. Thus $S$ is a basis for $\mathbb R\langle S\rangle$.

Let $V$ and $W$ be finite-dimensional real vector spaces and let $\mathcal R$ be the subspace of $\mathbb R\langle V\times W\rangle$ spanned by elements of the form \begin{align}a\left(v,w\right)&-\left(av,w\right),\\a\left(v,w\right)&-\left(v,aw\right),\\\left(v,w\right)+\left(v',w\right)&-\left(v+v',w\right),\\\left(v,w\right)+\left(v,w'\right)&-\left(v,w+w'\right).\end{align} The tensor product of $V$ and $W$, denoted $V\otimes W$, is $\mathbb R\langle V\times W\rangle/\mathcal R$ and the equivalence class of an element $\left(v,w\right)$ in $V\otimes W$, denoted $v\otimes w$, is the tensor product of $v$ and $w$.

Proposition (Characteristic Property of Tensor Products). Let $V$ and $W$ be finite-dimensional real vector spaces. If $A:V\times W\to X$ is bilinear into a vector space $X$, then there is a unique linear map $\tilde A:V\otimes W\to X$ such that the following diagram commutes:

where $\pi\left(v,w\right)=v\otimes w$.