Stacking powers: the tensor algebra
Write $V^{\otimes k} = V \otimes \cdots \otimes V$ ($k$ factors), with $V^{\otimes 0} = R$ and $V^{\otimes 1} = V$. The tensor algebra is the direct sum $T(V) = \bigoplus_{k \ge 0} V^{\otimes k}$, with multiplication given by concatenation: $(v_1 \otimes \cdots \otimes v_p)\cdot(w_1 \otimes \cdots \otimes w_q) = v_1 \otimes \cdots \otimes v_p \otimes w_1 \otimes \cdots \otimes w_q$. This makes $T(V)$ a graded algebra — graded by tensor degree — and in fact the free associative algebra on $V$: any linear map $V \to A$ into an associative algebra extends uniquely to an algebra map $T(V) \to A$. That is once again a universal property.
$T(V)$ is huge and noncommutative — $v \otimes w \ne w \otimes v$ in general. The exterior and symmetric algebras are the two most important quotients of $T(V)$, obtained by imposing one extra relation. Think of $T(V)$ as the raw material and the next algebras as what you get by demanding a sign rule (exterior) or commutativity (symmetric).
Imposing $v \wedge v = 0$
The exterior algebra is $\Lambda(V) = T(V) / I$, where $I$ is the two-sided ideal generated by all $v \otimes v$. The induced product is the wedge product $\wedge$, and the defining relation $v \wedge v = 0$ forces antisymmetry: $0 = (v+w)\wedge(v+w) = v\wedge w + w\wedge v$, so $v \wedge w = -\,w \wedge v$. More generally swapping any two adjacent factors flips the sign — wedging is the algebra of alternating, i.e. sign-flipping, multilinear maps.
Dimensions of the exterior powers. If dim V = n, then
dim Lambda^k(V) = C(n, k) (binomial), and total dim Lambda(V) = 2^n.
Basis of Lambda^k(V): { e_{i_1} ^ ... ^ e_{i_k} : i_1 < i_2 < ... < i_k }.
Why 'i_1 < ... < i_k'? Any repeat gives 0 (since e_i ^ e_i = 0), and any
reordering only changes the sign, so strictly increasing indices list each
basis element exactly once.
Worked example, n = 3, basis e_1,e_2,e_3:
Lambda^0 : 1 (dim 1)
Lambda^1 : e_1, e_2, e_3 (dim 3)
Lambda^2 : e_1^e_2, e_1^e_3, e_2^e_3 (dim 3)
Lambda^3 : e_1^e_2^e_3 (dim 1) <- the TOP power, one-dimensional
Compute (e_1 + 2e_2) ^ (e_2 - e_3):
= e_1^e_2 - e_1^e_3 + 2 e_2^e_2 - 2 e_2^e_3
= e_1^e_2 - e_1^e_3 - 2 e_2^e_3 (the 2 e_2^e_2 term vanishes).Wedges measure oriented volume
Geometrically, $v_1 \wedge \cdots \wedge v_k$ represents the oriented $k$-volume of the parallelepiped spanned by the $v_i$. Antisymmetry encodes orientation (swap two edges, flip the sign) and degeneracy (two equal edges give zero volume). A set of vectors is linearly independent iff its wedge is nonzero — wedging is a clean, basis-free independence detector.