Proof of correlation function formula in quantum field theory

I am trying to prove the following formula used in QFT:

$$\langle\Omega|T\{\Phi(x_1)\dots\Phi(x_n)\}|\Omega\rangle = \frac{\langle 0|T\{\Phi_I(x_1)\dots\Phi_I(x_n)S\}| 0 \rangle}{\langle 0|S| 0 \rangle}$$

where operators on LHS are in Heisenberg picture, on RHS they are on Interaction picture; and $S=U(+\infty, -\infty)=T\big\{exp\big(-i\int_{-\infty}^{+\infty} dt' H_I(t')\big)\big\}$. I am stuck at a certain step. This is what I'm trying to do:

$$\langle\Omega|T\{\Phi(x_1)\dots\Phi(x_n)\}|\Omega\rangle = \langle\Omega|T\{U^\dagger(t_1,t_0)\Phi_I(x_1)U(t_1,t_0)\dots U^\dagger(t_n,t_0)\Phi_I(x_n)U(t_n,t_0)\}|\Omega\rangle = \langle\Omega|T\{\Phi_I(x_1)\dots \Phi_I(x_n)\}|\Omega\rangle$$

As operators commute inside the $T\{\}$. Now I insert two identities:

$$ = \langle\Omega|T\{U^\dagger(T,t_0)U(T,t_0)\Phi_I(x_1)\dots \Phi_I(x_n)U^\dagger(-T,t_0)U(-T,t_0)\}|\Omega\rangle = \langle\Omega|T\{U^\dagger(T,t_0)\Phi_I(x_1)\dots \Phi_I(x_n)U(T,-T)U(-T,t_0)\}|\Omega\rangle = \langle\Omega|U^\dagger(T,t_0)T\{\Phi_I(x_1)\dots \Phi_I(x_n)U(T,-T)U(-T,t_0)\}|\Omega\rangle$$

Where in the last step, I've taken the arbitrary $t_0> t_1,\dots,t_n$, so that $U^\dagger(T,t_0)$ has only operators which are at later times than the rest, and so can be taken outside the time ordered part to the left, and it'll act on the $\langle\Omega|$

Now, I want to do the same with the $U(-T,t_0)$, taking it outside, on the right so that it may act on the $|\Omega\rangle$. However, the interval $(t_0, -T)$, includes times that are later than some of the $t_1, \dots, t_n$, and so parts of it won't be on the right when time ordered, so can't be taken out?

What can I do? Thanks!

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Here is an alternative proof:

As shown here,

$$|\Omega\rangle=\frac{e^{iE_\Omega(\frac{T}{2}+t_0)}}{\langle \Omega | 0 \rangle} U(t_0, -\frac{T}{2})|0\rangle$$

and

$$\langle\Omega|=\frac{e^{iE_\Omega(\frac{T}{2}-t_0)}}{\langle 0 | \Omega \rangle} \langle 0 |U(\frac{T}{2}, t_0)$$

where $T \rightarrow \infty(1-i\epsilon)$.Then

$$\langle\Omega|T\{\Phi(x_1)\dots\Phi(x_n)\}|\Omega\rangle = \frac{e^{iE_\Omega T}}{|\langle 0 | \Omega \rangle|^2}\langle0|U(\frac{T}{2}, t_0)T\{U^\dagger(t_1,t_0)\Phi_I(x_1)U(t_1,t_0)\\ \dots U^\dagger(t_n,t_0)\Phi_I(x_n)U(t_n,t_0)\}U(t_0, -\frac{T}{2})|0\rangle$$

However, I'm stuck at the same point. Not being able to put both $U$ operators inside the time ordering for any particular choice of $t_0$.