Simplification of nested time-ordered products I'm trying to progress towards understanding, and perhaps finding a proof for, the "nested" Wick's theorem for time-ordered products $T\{ \ldots \}$ alluded to in part (II) of this answer.
Assuming bosonic operators for now, I've noticed that
$$T\{ T\{ A(t_1)B(t_2) \} T\{ C(t_3)D(t_4) \} \} \equiv T\{ A(t_1)B(t_2)C(t_3)D(t_4) \}\tag{1}$$
through brute force calculation. Is the natural generalisation of this,
$$T\{ T\{ \ldots_1 \} \ldots_2 T\{ \ldots_3 \} \ldots_4 ~~~\ldots~~~ T\{ \ldots_{n-1} \} \ldots_n \} \equiv T\{ \ldots_1 \ldots_2 ~~~\ldots~~~ \ldots_n \},\tag{2}$$
also true?
 A: *

*Let us first define a $n$-ary Heaviside step function:
$$\begin{align} \theta&(t_1\geq t_2\geq\ldots \geq t_n)\cr
~:=~&\left\{\begin{array}{rl} 0&\text{if ineq. is violated}, \cr
\frac{1}{m_1!\ldots m_r!}&\text{if ineq. holds and there are $r$ sets of equal} \cr
&\text{times with multiplicities } m_1, \ldots, m_r.   \end{array} \right.\end{align} \tag{A}$$
It satisfies
$$ \sum_{\pi\in S_n} \theta(t_{\pi(1)}\geq t_{\pi(2)}\geq\ldots \geq t_{\pi(n)})~=~1. \tag{B}$$


*Next define time-ordering $T$ for Grassmann-even$^1$ operators as
$$ T(A_1 \ldots A_n)~:=~\sum_{\pi\in S_n} \theta\left( t(A_{\pi(1)})\geq \ldots \geq t(A_{\pi(n)})\right) A_{\pi(1)} \ldots A_{\pi(n)}. \tag{C}$$
Time-ordering $T$ is a multi-linear map and it may be viewed as a symmetric operad
$$T(A_1 \ldots A_n)~=~T(A_{\pi(1)} \ldots A_{\pi(n)}), \qquad \pi~\in~ S_n.\tag{D} $$
It satisfies a generalized idempotency
$$\begin{align}T&\left(T(A_1 \ldots A_r)T(B_1 \ldots B_s)\ldots Z_1 \ldots Z_u \right)\cr
~\stackrel{(C)}{=}~~&\sum_{\pi\in S_r} \theta\left( t(A_{\pi(1)})\geq \ldots \geq t(A_{\pi(r)})\right)\cr
&\sum_{\sigma\in S_s} \theta\left( t(B_{\sigma(1)})\geq \ldots \geq t(B_{\sigma(s)})\right)\ldots\cr 
&T\left(A_{\pi(1)} \ldots A_{\pi(r)}B_{\sigma(1)} \ldots B_{\sigma(s)}\ldots Z_1 \ldots Z_u \right)\cr
~\stackrel{(B)+(D)}{=}&T(A_1 \ldots A_r B_1 \ldots B_s \ldots Z_1 \ldots Z_u )
,\end{align}\tag{E} $$
which is OP's eq. (2).
--
$^1$ There is as straightforward generalization to Grassmann-graded operators.
