# Form of the interaction representation time evolution operator

I am a bit confused about the interaction representation picture. Consider the time independent Hamiltonian $$H = H_0 + V$$. My question concerns the interaction representation time evolution operator: $$U_I(t) = U_0^\dagger(t)U(t) = e^{iH_0t}e^{-i(H_0 + V)t}.$$ On the one hand we have that the interaction representation state satisfies: $$|\psi_I(t)\rangle = U_0^\dagger(t)|\psi_S(t)\rangle = U_I(t)|\psi_S(0)\rangle = U_I(t)|\psi_I(0)\rangle.$$ On the other hand, it also be defined as the solution to the differential equation: $$\frac{d}{dt}|\psi_I(t)\rangle = -iV_I(t)|\psi_I(t)\rangle \quad |\psi_I(0)\rangle = |\psi_S(0)\rangle,$$ where $$V_I(t) = U_0^\dagger(t)VU_0(t)$$, meaning that $$U_I(t) = \operatorname{T}\exp \left(-i\int_0^t V_I(\tau)d\tau\right).$$ My question is: am I right to conclude from this that: $$e^{iH_0t}e^{-i(H_0 + V)t} = \operatorname{T}\exp \left(-i\int_0^t V_I(\tau)d\tau\right)?$$ If it is true, can it be shown directly from the definition, without reference to the Schrodinger equation? My guess is that applying the BCH formula to the left hand side would give you the time ordered exponential somehow, but I am struggling to see where the connection is. I'm not asking for a proof necessarily, but rather just how one would in principle show that this is true.

In fact the reference to the state vector is not essential for the derivation. Directly take derivative of $$U_I(t)$$ with respect to $$t$$:
$$\displaystyle \frac{d}{dt}U_I(t)=iH_0e^{iH_0t}e^{-i(H_0+V)t}-ie^{iH_0t}(H_0+V)e^{-i(H_0+V)t} \\ = -i e^{iH_0t}Ve^{-iH_0t}e^{iH_0t}e^{-i(H_0+V)t}\\ =-iV_I(t)U_I(t)$$
$$\displaystyle U_I(t)=\mathbf{1}-i\int_0^td\tau\, V_I(\tau)U_I(\tau).$$