Let $\Delta\_N := \{0 \leq \tau\_1 \leq \dots \leq \tau\_N \leq 1\}$ be the $N$-simplex. Let $a\_j: [0, 1] \rightarrow \mathcal{A}$, $j=1, \dots, N$ be continuous functions with values in some (non-commutative) finite-dimensional Banach algebra (e.g. some matrix algebra). Consider the $\mathcal{A}$-valued integral
$$
 \mathrm{Int} = \sum\_{\sigma \in \mathcal{S}\_N} \mathrm{sgn}(\sigma) \int\_{\Delta\_N} a\_{\sigma\_N}(\tau\_N) \cdots a\_{\sigma\_1}(\tau\_1)\, \mathrm{d} \tau\_1 \cdots \mathrm{d} \tau\_N,
$$
where $\mathcal{S}\_N$ denotes the group of permutations of the numbers $\{1, \dots, N\}$.
Suppose that the index set $\{1, \dots, N\}$ splits as the disjoint union of non-empty sets $I\_1, \dots, I\_k$ such that we elements from different sets $I\_u$ anti-commute pairwise, i.e. for $i \in I\_u$, $j \in I\_v$, one has
$$ a\_i(t)a\_j(s) = -a\_j(s)a\_i(t), ~~~~~~~~ \forall s, t \in [0, 1].$$
Clearly, if $k=N$, i.e. each set has only one element so that all the $a\_j$ anti-commute, then
$$\mathrm{Int} = \prod\_{j=1}^N \int\_{\Delta\_1}a\_j(\tau) \mathrm{d}\tau.$$
In the general case, we guess that one has
$$ \mathrm{Int} = \prod\_{j=1}^k \sum\_{\sigma \in \mathcal{S}\_{I\_j}} \mathrm{sgn}(\sigma) \int\_{\Delta\_{|I\_j|}} \prod\_{i\in I\_j} a\_{\sigma\_i}(\tau\_i) \mathrm{d}\tau\_i,$$
where $\mathcal{S}\_{I\_j}$ denotes the group of permutations of the set $I\_j$ and the latter product is run through from the largest to smallest number $i \in I\_j$.


We were unable to prove this in full generality so far, because we didn't find a way to efficiently handle the combinatorics. Does anybody have hints how to tackle this problem or probably people already looked at this so that there is a reference?



Combinatorics of integral over simplices
For a $\sigma\in S\_N$, let $\Delta\_N^{\sigma}=\{(\tau\_1, \ldots, \tau\_N)\mid 0\le \tau\_{\sigma\_1}\le\cdots\le\tau\_{\sigma\_N}\le 1\}$. Clearly, $\Delta\_N^{\sigma}$ is a space homeomorphic to $\Delta\_N$, and there is a decomposition $$I^N=\bigcup\_{\sigma\in S\_N} \Delta\_N^{\sigma}.$$
The points of $I^N$ that belong to more than one of the $\Delta\_N^{\sigma}$s form a subset of measure zero (it is the set of $N$-tuples that are not all distinct). In what follows, when I say something is a function on $I^N$, it is only required to be defined unambiguously outside this set of measure zero. 


With this caveat in mind, let $\tilde\sigma\colon I^N \to S\_N$ be the "function" that assigns to an $n$-tuple $(t\_1, \ldots, t\_N)$ the permutation $\sigma$ for which $t\_{\sigma\_1}\le\cdots \le t\_{\sigma\_N}$. Then your Int can be rewritten as an integral over $I^N$, as follows
$${\mbox{Int}}= \int\_{I^N} {\mbox {sgn}}(\tilde\sigma) \cdot a\_{{\tilde \sigma}\_N}(\tau\_{{\tilde \sigma}\_N})\cdots a\_{{\tilde \sigma}\_1}(\tau\_{{\tilde \sigma}\_1})\, d\tau\_1\cdots d\tau\_N.$$ Let us denote the integrand function by $f\_N(\tau\_1, \ldots, \tau\_N)$. 


Now let us suppose that $I\_1=\{1, \ldots, i\_1\}$, $I\_2=\{i\_1+1,\ldots i\_1+i\_2\}$ and so on up to $I\_k$. There is a canonical identification $I^N\cong I^{I\_1}\times \cdots \times I^{I\_k}$. Moreover, your hypothesis about anti-commuting elements is equivalent to saying that the function $f\_N$ splits as a product $$f\_N(\tau\_1, \ldots, \tau\_N)= f\_{I\_1}(\tau\_1, \ldots, \tau\_{i\_1})\cdot f\_{I\_2}(\tau\_{i\_1+ 1},\cdots, \tau\_{i\_1+ i\_2})\cdots f\_{I\_k}(\cdots).$$


It follows that
$$\int\_{I^N}f\_N = \int\_{I^{I\_1}}f\_{I\_1}\cdots \int\_{I^{I\_k}}f\_{I\_k}.$$
I think this is equivalent to the formula you are asking.


