Why should the Fermi level of a n-doped semiconductor be below the one of a p-doped? In a pn-junction, the difference in Fermi level between the p doped and the n doped regions causes the apparition of a built-in electric field at equilibrium. This electric field goes from the n to the p, (so the positive carriers, for example, would not feel anymore the Coulomb attraction from the ionized atom donors), meaning that the Fermi level of the n doped region is below the one of the p doped, but I don't see any elementary argument explaining that.
 A: Before the p-doped and n-doped materials are joined, maybe we can think that their conduction and valence bands are aligned (although that's probably a dubious  assumption). We know that when they join the Fermi-levels must be flat so we need to lower the n-type material down in energy. We lower the n side because electrons row down hill, that is to say we are minimising their energy. Or you could say we move the p-side up because hole roll up hill. The result is the same: the n-side is lower than the p-side. 
After the charge has equilibrated, the end result is the bands bend to accomodate the flat Fermi level.
 
A: In equilibrium the Fermi level (or chemical potential ) must not change across the junction - this is exactly the thermodynamic condition of equilibrium.
When n doped semiconductor is set into contact with the p-doped semiconductor it is true that an intrinsic field is set up at the junction and this is exactly what serves to align the chemical potentials. Thus the Fermi level of n-doped semiconductor should be exactly the same of the p-doped semiconductor that it is on contact with (in equilibrium conditions).
A: The reason is an n doped material has additional atoms (donor atoms) added near the conduction band. They provide electrons which can easily go to the conduction band, as the energy gap between the donor level and the conduction band is low. There are now more levels above the Fermi level, thus skewing the distribution of electrons higher. This is seen by the Fermi level moving up.
In the p doped case, Boron or other group 3 elements are added, more energy levels closer to the valence band are added. They can take an electron from the valence band, creating a hole in the valence band. In this case, there are now more energy levels below the Fermi level, so the probability of the electron being in a lower level is higher. This skews the Fermi level down, closer to the conduction band. 
Why? Because the Fermi level has a 50% chance of being filled at 0K. That is its definition. 
Fermi function, $f$ is $$f = \frac{1}{1+e^{(E-E_f)/k_B T}} $$
When $E=E_f$, $f = 0.5$. The probability is 0.5.
If acceptor levels are added (p dopants), you might think the Fermi level needs to go up, but it goes down, closer to the acceptor levels, because the number of STATES increases, not the number of electrons.
If donor levels are added, then electrons do get added. However, in this case the valence band is assumed to be full, and adding more states/energy levels above the Fermi level shifts it up. For every electron you add, you add two states, thus the same effect is seen as the acceptor levels case, but not as powerful. 
