A step in the derivation of the unitary time-evolution operator (time-ordered exponential) I'm reading introductory quantum field theory, from the book Relativistic Quantum Physics by T. Ohlsson. In the derivation of the unitary time-evolution operator, chapter 11.2, there is an equation that I get stuck on. Here it is, with a little bit of context:

... In the case of fermions, the shifted interaction Hamiltonian $H'_I$ always includes a product of an even number of fermions. Thus, for the ${H'_I}^2$ term, we can symmetrize the integrals in the following way
$$\int_{t'}^t dt_1 \int_{t'}^{t_1} dt_2 T\left(H'_I(t_1)H'_I(t_2)\right) = \int_{t'}^t dt_2 \int_{t'}^{t_2} dt_1 T\left(H'_I(t_1)H'_I(t_2)\right)\\ = \frac{1}{2} \int_{t'}^t dt_1 \int_{t'}^{t} dt_2 T\left(H'_I(t_1)H'_I(t_2)\right).$$

I have been trying to verify this by using the definition of the time ordered product for fermionic fields,
$$T\left(\psi_1(x_1)\psi_2(x_2)\right) = \theta(t_1 - t_2)\psi_1(x_1)\psi_2(x_2) - \theta(t_2 - t_1)\psi_2(x_2)\psi_1(x_1),$$
but I'm not really getting anywhere. The way it is introduced makes it seem like it should be obvious, and maybe it is, but I'm not getting it, so I would greatly appreciate if anyone could show me how the quoted equalities are derived.
 A: $$\int_{t'}^t dt_1 \int_{t'}^{t_1} dt_2 \ T\{H'_I(t_1)H'_I(t_2)\} \overset{(1)}{=} \int_{t'}^t dt_2 \int_{t'}^{t_2} \ dt_1 T\{H'_I(t_1)H'_I(t_2)\} \\ \overset{(2)}{=}  \frac{1}{2} \int_{t'}^t dt_1 \int_{t'}^{t} dt_2 T\left(H'_I(t_1)H'_I(t_2)\right).$$
Where in the first equality we use time ordering symmetry, and in the second we use an integration trick which, at second order, is summed up well in this diagram:

[Peskin and Schroeder, An Introduction to Quantum Field Theory, Figure 4.1]
and it's not the greatest leap of faith to see how this holds at order $N$, with the time-ordered product of $N$ Hamiltonian operators.
Now in P&S, this expression is derived for Hamiltonians of bosonic fields, since it relies on the symmetry of the time-ordered expression about $t_1=t_2...=t_n$. For fermionic fields, recall that the time ordering operation is not a blanket command to be able to shuffle fields within it willy-nilly - one must remember to include minus signs when a Grassman-odd fermion is anticommuted past another. However, when the interaction part of the Hamiltonian contains products of an even number of fermionic fields, then you will pick up factors of $(-1)^{2n}=1$ for moving $H'_I(t_l)$ past $H'_I(t_m)$, so this symmetry is retained.
