Why is the bar complex free?

Question

I have seen in multiple places now, that when working over a field $\mathbb{k}$, the bar complex of a $\mathbb{k}$-algebra A, $$ \cdots \rightarrow A^{\otimes 3} \rightarrow A^{\otimes 2},$$ is a free $A^e$-module resolution of A. I know that it is projective, see Weibel for instance, but I haven't found a source that shows this is a free resolution. I personally cannot see why this sucker is free. Any thoughts?

Answer 1

Show that the free $A^e$-module on a vector space $V$ (equivalently, on a basis of $V$) is $A \otimes V \otimes A$. Now set $V = A^{\otimes (n-2)}$.