quantum mechanics - Why does time evolution operator have the form $U(t) = e^{-itH}$?

Let's denote by $|\psi(t)\rangle$ some wavefunction at time $t$. Then let's define the time evolution operator $U(t_1,t_2)$ through

$$U(t_2,t_1) |\psi(t_1)\rangle = |\psi(t_2)\rangle \tag{1}$$

and Hamiltonian $H$ through

$$H |\psi(t)\rangle = i\frac{\partial}{\partial t} |\psi(t)\rangle. \tag{2}$$

(We have set $\hbar = 1$.)

Question.

What is the relation between $U$ and $H$, given that $H$ doesn't depend explicitly on $t$?

Attempts.

It is well-known that the answer is $U(t_2,t_1) = c \cdot e^{-i(t_2-t_1)H}$ for some scalar $c$.

Some sources postulate it follows from operator equation $HU = i \frac{\partial}{\partial t} U$ but we cannot integrate it like we could an ordinary differential equation $f(x)g(x) = \frac{\partial}{\partial x} g(x)$ (at least I don't know how to integrate operators!).

Then we could also "guess" the solution $U(t_2,t_1) = c \cdot e^{-i(t_2-t_1)H}$ and verify by inserting into $(1)$, but then we still need to show that this is the general solution, i.e., $e^{-i(t_2-t_1)H}$ spans the set of all solutions. Also this approach doesn't enlighten as to why we chose that particular guess.

Answer

For time-independent Hamiltonians, $U(t) = \mathrm{e}^{\mathrm{i}Ht}$ follows from Stone's theorem on one-parameter unitary groups, since the Schrödinger equation

$$H\psi = \mathrm{i}\partial_t \psi$$ is just the statement that $H$ is the infinitesimal generator of a one-parameter group parameterized by the time $t$.

For time-dependent Hamiltonians $H(t) = H_0 + V(t)$, the time evolution actually is dependent on the start- and end-points, and the Schrödinger equation is iteratively solved by a Dyson series in the interaction picture, whose schematic form is

$$U(t_1,t_2) = \mathcal{T}\exp(\int_{t_1}^{t_2}\mathrm{e}^{\mathrm{i}H_0 t}V(t)\mathrm{e}^{-\mathrm{i}H_0t}\mathrm{d}t)$$ in the Schrödinger picture, and one obtains it from the evolution equation in the interaction picture, which is the Tomonaga-Schwinger equation $$\mathrm{i}\partial_t U_I(t,t_0)\psi_I(t_0)\lvert_{t=t_1} = V(t_1)U_I(t_1,t_0)\psi_I(t_0)$$ and iterating the solution $$U(t_1,t_0) = 1 - \mathrm{i}\int_{t_0}^{t_1}U(t,t_0)\mathrm{d}t$$