Simplex integral in connection with time ordered exponential In Quantum Mechanics, one often defines the time ordered exponential like e.g. here.
Now my question is how the factor of $N!$ arises. I know the simplex volume as the following integral: 
\begin{equation}
\int_{t_0}^t dt_1 \int_{t_1}^{t} dt_2 \cdots \int_{t_{N-1}}^t dt_N = \frac{(t-t_0)^N}{N!}
\end{equation}
However, I would like to know how to obtain the identity
\begin{equation}
\int_{t_0}^t dt_1 \int_{t_1}^{t} dt_2 \cdots \int_{t_{N-1}}^t dt_N~f(t_1)\cdots f(t_N) = \frac{1}{N!} \int_{t_0}^t dt_1\cdots \int_{t_{0}}^{t} dt_N~\mathbb{T}~(f(t_1)\cdots f(t_N))
\end{equation}
where $\mathbb{T}$ is the time-ordering operator that acts as follows:
\begin{equation}
\mathbb{T}~(f(t_1)\cdots f(t_m))=f(t_{\pi(N)})\cdots f(t_{\pi(1)})\qquad\text{with}\qquad t_{\pi(N)}<\cdots < t_{\pi(1)}
\end{equation}
Thanks.
 A: Let's take the case where the $t_j$ are real numbers, the function $f$
is real valued, and $t_0=0$. The integral on the right hand side is over an
$N$-cube, which is a union of $N!$ many $N$-simplices. There is one
 simplex for
each permutation $\pi$ of $123\cdots N$, defined by
$t\ge t_{\pi(N)}\ge\cdots\ge t_{\pi(1)}\ge 0$.
For example with $N=2$ the simplices
$0\le t_1 \le t_2 \le 1$ and $0\le t_2\le t_1\le 1$ subdivide the unit square.
The integral on the right hand side is equal to the sum of
integrals over the simplices. In the summand corresponding
to a permutation $\pi$, make the change of variables
$$
 s_j = t_{\pi(j)}.
$$
Then $s_N\ge\cdots\ge s_1$, so the mapping from $s$ to $t$
takes a standard simplex to the one indexed by $\pi$.
The matrix derivative of this transformation has determinant
$\pm 1$, the signature of $\pi$. By the change of variables theorem
for multiple integrals then
$$
 \int_{t\ge t_{\pi(N)}\ge\cdots\ge t_{\pi(1)}\ge 0}
     f(t_{\pi(1)})\cdots f(t_{\pi(N)}) d^N t
 =
 \int_{t\ge s_N\ge\cdots\ge s_1\ge 0}f(s_1)\cdots f(s_N)\,d^N s.
$$
[By the way this shows that you have written the time-ordering
in the wrong order.]
This last integral is equal to the iterated integral on the left
of your equation. Since each of the $N!$ integrals on the right
is equal to the one integral on the left, it is
necessary to divide the sum on the right by $N!$.
Your reference in Wikipedia allows $t$ in a ``real or complex field''
and $f(t_j)$ in an algebra over that field. I don't know what
the integrals mean unless the $t_j$ are real, but you might be
able to extend the argument to values in finite dimensional
algebras by expanding the values $f(t_j)$ in terms of a basis, with
coefficients that are real valued functions of $t$.
