Here are two questions related to the following form $$U(t,t_0)=1+(\frac{-i}{\hbar})\int_{t_0}^tdt'\,H^I_1(t')U(t',t_0) \tag{1}$$ This is known as the iterative form of solution to the differential equation for operators: $$i\hbar\frac{\partial}{\partial t}U(t,t_0)=H^I_1(t)U(t,t_0) $$ where $H^I_1(t)$ is known as the Hamiltonian in the interaction picture. $U(t_0,t_0)=1$ ($U, H, 1$ are operators).
Now my question is, suppose we know the following property of $U(t,t_0):$ $$U(t,t_1)U(t_1,t_0)=U(t,t_0) $$ for arbitrary $t,\,t_1$ and $t_0$. Or $U(t,t_0)$ is a flow. Can we use the flow property together with $(1)$ to obtain the solution: $$U(t,t_0)=1+(\frac{-i}{\hbar})\int^t_{t_0}dt_1H^I_1(t_1)+(\frac{-i}{\hbar})^2\int^t_{t_0}dt_1\int^{t_1}_{t_0}dt_2 H^I_1(t_1)H^I_1(t_2)+(\frac{-i}{\hbar})^3\int^t_{t_0}dt_1\int^{t_1}_{t_0}dt_2\int_{t_0}^{t_2}dt_3H^I_1(t_1)H^I_1(t_2)H^I_1(t_3)+\dots=\sum_{n=0}^{\infty}(\frac{-i}{\hbar})^n\int^t_{t_0}dt_1\int^{t_1}_{t_0}dt_2\dots\int^{t_{n-1}}_{t_0} dt_n H_1^I(t_1)\dots H_1^I(t_n) $$
I tried to split $U(t,t_0)$ in terms of $U(t,t_1)U(t_1,t_2)U(t_2,t_3)\dots U(t_n,t_0)$ and let $n\to \infty$, but I failed because I know $HU$ do not commute ingeneral. Can someone give me a hint?
$$U(t,t_0)=1-\frac{i}{\hbar}\int_{t_0}^tdt_{1}\,H^{I}(t_{1})U(t_{1},t_0) \tag{1}$$ Insert the equation into itself to give $$U(t,t_0)=1-\frac{i}{\hbar}\int_{t_0}^{t}dt_{1}\,H^{I}(t_{1}) \tag{1}-\frac{1}{\hbar^{2}}\int_{t_{0}}^{t}dt_{1}\int_{t_{0}}^{t_{1}}dt_{2}H^{I}(t_{1})H^{I}(t_{2})U(t_{2}, t_{0})$$ Once more $$U(t,t_0)=1-\frac{i}{\hbar}\int_{t_0}^{t}dt_{1}\,H^{I}(t_{1}) \tag{1}-\frac{1}{\hbar^2}\int_{t_{0}}^{t}dt_{1}\int_{t_{0}}^{t_{1}}dt_{2}H^{I}(t_{1})H^{I}(t_{2})+$$ $$+\frac{i}{\hbar^{3}}\int_{t_{0}}^{t}dt_{1}\int_{t_{0}}^{t_{1}}dt_{2}\int_{t_{0}}^{t_{2}}dt_{3}H^{I}(t_{1})H^{I}(t_{2})H^{I}(t_{3})U(t_{3}, t_{0})$$ Continue inserting the equation into itself to get $$U(t, t_{0})=1+\sum_{k=1}^{\infty}\Big(-\frac{i}{\hbar}\Big)^{k}\int_{t_{0}}^{t}dt_{1}\int_{t_{0}}^{t_{1}}dt_{2}...\int_{t_{0}}^{t_{k-1}}dt_{k}\mathcal{T}[H^{I}(t_{1})...H^{I}(t_{k})]$$