I think a lot of people run into this issue. The way I think about it is the following:

Take your finite-dimensional vector space $V$ and form its tensor algebra $T(V)$. Define $\mathcal{J}$ to be the 2-sided ideal in $T(V)$ generated by elements of the form $v \otimes v$, and then define the exterior algebra to be $\Lambda(V) = T(V) / \mathcal{J}$. This exhibits the exterior algebra as a quotient of the tensor algebra.

The different conventions you see for the wedge product arise from different embeddings of the exterior algebra into the tensor algebra. Define on $V^{\otimes n}$ the map
$$ A_n (v_1 \otimes \dots \otimes v_n) = \frac{1}{n!} \sum_{\pi \in S_n} sgn(\pi) v_{\pi(1)} \otimes \dots \otimes v_{\pi(n)},
$$
(or possibly with $\pi^{-1}$ instead of $\pi$, although I guess it doesn't matter)
and then define on the tensor algebra the map
$$A = \bigoplus_{n=0}^{\infty} A_n.$$
Then you can show easily that $A_n^2 = A_n$ for all $n$, so that $A$ is a projection.

The point is that $\mathcal{J} = \mathrm{ker} (A)$, so that you can identify the quotient $\Lambda(V)$ with $\mathrm{im} A$, i.e. we have now embedded the exterior algebra as a subspace of the tensor algebra. This is where the two conventions differ. I have defined $A_n$ with a $\frac{1}{n!}$ in front, but some don't do so. Of course, this doesn't change the kernel of the map, but it does change the embedding of the exterior algebra into the tensor algebra.

The important point is that $A$ is not an algebra map of $T(V)$ to itself, so the embedding $\Lambda(V) \to T(V)$ is not an embedding of algebras. Now you ask how to describe the exterior product in terms of the product in $T(V)$. Take $\alpha \in \Lambda^k(V)$ and $\beta \in \Lambda^l(V)$ with representatives $\tilde{\alpha} \in \mathrm{im}(A_k)$ and $\tilde{\beta} \in \mathrm{im}(A_l)$, respectively. Then $A_{k+l}(\tilde{\alpha} \otimes \tilde{\beta})$ is the representative of $\alpha \wedge \beta$ that you're looking for.

Essentially, it boils down to whether or not you put the $\frac{1}{n!}$ in front of your alternating map or not.

quotientof tensor algebra by the relation $\alpha\wedge \alpha=0$. Forcing it inside the tensor algebra is ugly and unnatural. – Donu Arapura Feb 4 '11 at 18:54