= Solution
Start with the <Schrödinger equation> $i\hbar\partial_t|\psi\rangle=(H_0+V)|\psi\rangle$ and differentiate the definition of the <interaction picture> state:
$$
\begin{aligned}
i\hbar\frac{d}{dt}\left(e^{iH_0t/\hbar}|\psi\rangle\right)
&=-H_0e^{iH_0t/\hbar}|\psi\rangle
+e^{iH_0t/\hbar}(H_0+V)|\psi\rangle\\
&=e^{iH_0t/\hbar}Ve^{-iH_0t/\hbar}|\psi_I\rangle.
\end{aligned}
$$
The free <Hamiltonian operator> terms cancel because $H_0$ commutes with its own <matrix exponential>. Therefore
$$
\boxed{i\hbar\frac{d}{dt}|\psi_I(t)\rangle=V_I(t)|\psi_I(t)\rangle,
\qquad V_I(t)=e^{iH_0t/\hbar}V(t)e^{-iH_0t/\hbar}.}
$$
This <interaction picture> transformation is exact.
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