This is a problem from quantum mechanics, but purely mathematical (I've set $\hbar = 1$):
Let $\psi(\cdot, t) \in L_2(\mathbb{C}^3)$ for all $t\geq 0$ and $\hat{H}$ a bounded, hermitian linear operator on $ L_2(\mathbb{C}^3)$, define $$\hat{U} := \exp\bigl(-i \hat{H} t\bigr)\,.$$ If we set $$\psi(x, t) = \hat{U}(t)\psi(x,0)$$ then $\psi(x,t)$ is a solution of the differential equation (Schrödinger equation) $$i \frac{\partial}{\partial t} \psi(x, t) = \hat{H} \psi(x, t)\, ,$$ as one easily shows: $$i \frac{\partial}{\partial t}\psi(x, t) = i \frac{\partial}{\partial t} \hat{U}(t) \psi(x, 0) = \hat{H}\hat{U}(t)\psi(x, 0) = \hat{H} \psi(x, t)\, .$$
Question: If we assume that the exponential of a linear operator is defined by the power series: $$\exp(A)\psi = \sum_{k=0}^\infty \frac{1}{k!} A^k\psi\, ,$$ how can we prove that indeed the following step $$i \frac{\partial}{\partial t} \hat{U}(t) \psi(x, 0) = \hat{H}\hat{U}(t)\psi(x, 0)$$ is correct? After all, the partial derivative is "tied" to the function $\hat{U}(t) \psi(x, 0)$.
If your Hamiltonian were bounded then indeed you could define its exponential via a power series (Although mostly in quantum mechanics the Hamiltonian contains position and momentum operators which are unbounded. Then you would have to use arguments of semigroup theory or functional calculus to solve the Schrödinger equation - which I am not very familiar with)
Anyhow, the power series you wrote down is not the exponential of a bounded operator, you missed a factor there: $$ \exp(A)\psi := \sum_{k=0}^\infty \frac{1}{k!}A^k\psi . $$ I would solve Schrödinger's equation in the Hilbert space $$ i\frac{\partial}{\partial t} \left|\psi(t)\right\rangle= H\left|\psi(t)\right\rangle $$ rather than in terms of wave functions (because of the top of my head I am not entirely sure why $\left\langle x\left|\frac{\partial}{\partial t}\right.\psi(t)\right \rangle = \frac{\partial}{\partial t}\left\langle x\left|\right.\psi(t)\right \rangle = \frac{\partial}{\partial t}\psi(x,t)$).
The solution to this equation in the Hilbert space is $$ \left|\psi(t)\right\rangle = U(t)\left|\psi(0)\right\rangle = \sum_{k=0}^\infty \frac{1}{k!}(-it)^kH^k\left|\psi(0)\right\rangle. $$
If you use that and accept that you can differentiate a power series term by term than you should be able to prove that in fact $$ i\frac{\partial}{\partial t} U(t)\left|\psi(0)\right\rangle = HU(t)\left|\psi(0)\right\rangle $$ is correct. Now to show the term by term differentiation for power series of (bounded) operators I think one can adapt the proof of the real analysis analogue.