You are correct. When the power supply is switched on the current through the inductor will be increasing (di/dt > 0). This will produce a transient voltage across the inductor that resists the applied voltage (a "back emf", following the passive sign convention for i and vL). Once the current reaches its maximum, di/dt = 0 and the voltage across the inductor is also zero. The circuit has reaches a steady state, or DC ("Doesn't Change").
However, another change will occur when the power supply is suddenly switched off. di/dt < 0, so the inductor assumes a negative voltage that tries to keep the current pumping. Both of these inductive voltage spikes can damage other parts of the circuit, particularly integrated circuits and transistors used for switching. This is why inductive loads (e.g. motors, solenoids, relays, etc.) are provided with flyback diodes which allow the current to keep flowing and dissipate the stored energy of the inductor's magnetic field:
simulate this circuit – Schematic created using CircuitLab