Why does voltage always lead current by 90 degrees in an inductor?

It's really that the current is the time-integral of the voltage, or the voltage is the derivative of the current. If the current is a sine, then the voltage is a cosine, since that's the derivative of a sine.

The way derivatives and integrals of sinusoids work, each is ¼ cycle, or 90°, phase shifted from the next.

The bottom line is the basic equation for an inductor and that equation applies in any electrical situation: -

$$V = L\dfrac{di}{dt}$$

So if the current is a sine wave, the differential of sine is cosine: -

Hence voltage leads current by 90 degrees. But remember this only applies to AC signal analysis. For instance if you applied a step voltage across an inductor the current rises linearly with time because: -

$$\dfrac{di}{dt} = \dfrac{V}{L}$$

The basic equation describes both AC and transient events.

Also, an ideal inductor with jwL has a positive imaginary part with no further real resistance. So the angle will turn 90°.