There is no delay. That would imply a time shift. (You can define delay in a certain way, e.g. the 50%-50% rise time, etc., but such a definition is arbitrary.)
It's simply that, for a constant voltage applied to an R+C, initially all the voltage (Vin - Vc(0)) appears across the resistor, so the capacitor current is maximal; in the next instant dt, the voltage approaches Vin an incremental amount dV = I/C dt, and so on. We integrate (substituting I = (Vin - Vc) / R), rearranging as needed to make it integrable, and obtain the exponential function.
There are no steps like 38%, 27, 16, etc., the process is completely continuous; we can define thresholds, which are proportional to the time constant (i.e., a ratio of ln(2) for example, where the proportion is 1), but which one we choose is arbitrary.
We also cannot consider the problem wholly stepwise, as we will get some kind of error if we do. If we perform the [Riemann] sum of ΔV = I/C Δt, what Δt shall we choose? Again it is arbitrary (the correct answer is lim Δt --> 0, and the sum becomes integral dt).
...This discussion may not be very helpful if you haven't taken calculus yet. May I recommend 3blue1brown's series on the subject?
Okay. Why emphasize the capacitor? Because the inductor is the same.
Everything an inductor does with current, a capacitor does with voltage.
Capacitor: I = C dV/dt
Inductor: V = L dI/dt
Just swap V and I.
We phrase it:
For a constant current applied to an R||L, initially all the current (Iin - IL(0)) drops through the resistor, so the inductor voltage is maximal; in the next instant dt, the current approaches Iin an incremental amount dI = V/L dt, and so on. We integrate (substituting V = (Iin - IL) * R), ...
Note that I used a series-parallel transformation here -- all voltages become currents and vice versa, so the source is actually a current source now, and the circuit is a current divider (everything in parallel), instead of a voltage divider (everything in series). The topology change may not be obvious, but it is easy to show that they are equivalent.
Interestingly (but by no accident), the result is precisely the same in the series circuit -- simply convert the parallel current source and resistor from Norton to Thevenin equivalent.
Tim