The $N$th-order identity of the time-ordered exponential in Quantum Mechanics 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!}=\frac{1}{N!}\int_{t_0}^t dt_1 \int_{t_0}^{t} dt_2 \cdots \int_{t_{0}}^t dt_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}
 A: We have that $\mathbb{T}~(f(t_1)\cdots f(t_N))=\mathbb{T}~(f(t_{\sigma(1)})\cdots f(t_{\sigma(N)}))$ for every permutation $\sigma$. We also know that if we have some function $F(t_1,...,t_N)$, then
$$
\int_{t_0}^t dt_1\cdots \int_{t_{0}}^{t} dt_N~F(t_{\sigma(1)},...,t_{\sigma(N)}) =\sum_\sigma\int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \cdots \int_{t_0}^{t_{N-1}} dt_N F(t_{\sigma(1)},...,t_{\sigma(N)}) 
$$
We thus have
$$
\begin{array}{rcl}
\int_{t_0}^t dt_1\cdots \int_{t_{0}}^{t} dt_N~\mathbb{T}~(f(t_1)\cdots f(t_N))&=&\sum_{\sigma}\int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \cdots \int_{t_0}^{t_{N-1}} dt_N~\mathbb{T}(f(t_{\sigma(1)})\cdots f(t_{\sigma(N)}))\\
&=&\sum_{\sigma}\int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \cdots \int_{t_0}^{t_{N-1}} dt_N~\mathbb{T}(f(t_1)\cdots f(t_{N}))\\
&=&\sum_{\sigma}\int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \cdots \int_{t_0}^{t_{N-1}} dt_N~f(t_1)\cdots f(t_{N})\\
&=&N!\int_{t_0}^t dt_1 \int_{t_0}^{t_1} dt_2 \cdots \int_{t_0}^{t_{N-1}} dt_N~f(t_1)\cdots f(t_{N})\\
\end{array}
$$
from which the identity immediately follows.
