Koszul differentials are not compatible with exterior algebras

Nobody has claimed this yet.

Let $K$ be the Koszul complex of an $n$-dimensional vector space $V$ and $E$ the exterior algebra on $n$ variables, then:
```math
K_i \simeq {\bigwedge}^i V \simeq E_i
```
The trouble is that the _order_ of the basis elements of $E_i$ is **not** the same as the order of the basis elements of $K_i$.

Here is the order on $E_2$, notice $e_0\wedge e_3 < e_1\wedge e_2$:
```m2
i21 : basis(2, E)
o21 = | e_0e_1 e_0e_2 e_0e_3 e_1e_2 e_1e_3 e_2e_3 |
```
Here is the differential $K_1 \gets K_2$:
```m2
i25 : K.dd_2

o25 = {1} | -x_1 -x_2 0 -x_3 0 0 |
{1} | x_0 0 -x_2 0 -x_3 0 |
{1} | 0 x_0 x_1 0 0 -x_3 |
{1} | 0 0 0 x_0 x_1 x_2 |

4 6
o25 : Matrix S <--- S
```
Notice the third and fourth columns in particular, sending the corresponding basis elements of $K_2$ to what would be `x_1*e_2 - x_2*e_1` and `x_0*e_3 - x_3*e_0`, which is the opposite of what $e_0\wedge e_3$ and $e_1\wedge e_2$ should be sent to.

Start by reproducing the `basis(2, E)` ordering and inspecting `K.dd_2` for the displayed four-dimensional example. Compare the differential signs with the exterior-algebra wedge convention; done means the corresponding Koszul and exterior-algebra bases produce compatible differentials, with a regression check for this example.

Written by the indexing model from the issue text.