Oh boy ...
There is the "ideal" situation, and then there is the "real" situation. The problem (and discrepancy between the two) is the heat transfer to and from the cylinder walls.
Without any heat transfer, compression of a gas follows a set of laws that have to do with the relationship between the heat capacity at constant volume (pressure varies in a fixed volume as you change the temperature), and the heat capacity at constant temperature (pressure stays the same, but the volume is allowed to expand). The physics behind this is very involved, but suffice it to say that for air at temperatures well above the critical point (the highest temperature at which the chemical compound can exist as a liquid) and pressures well below the critical point (the pressure which corresponds to above situation) that ratio is 1.4 and it is also 1.4 for other gases consisting of two atoms. Air actually contains a small amount of molecules consisting of a single atom, but also a small amount of molecules consisting of more than two atoms, but for this purpose they can be ignored.
If you want to know the final pressure P2 following compression to a factor n and starting with pressure P1 (note: all temperatures expressed in degrees above absolute zero, i.e. degrees K):
P2 = P1 * n ^ 1.4
The 1.4 is the ratio of specific heats.
With substitution of the ideal gas law, the temperature ratio can be extracted
T2 = T1 * n ^ 0.4
Note that this uses 0.4 and not 1.4 in the equation. If you substitute the ideal gas law into both sides of this equation and do some rearranging, you can see why this is the case.
So, for 16:1 compression ratio, T2 = T1 * 3.03
Suppose ambient temperature T1 is 20 C, i.e. 293 K, then T2 = 887 K = 614 C
When translating into reality, there are some troubles with this:
#1: There is heat transfer between the air and the cylinder walls and it is significant. If you are dealing with a stone-cold engine, that heat transfer is going only one way ... out of the air and into the cylinder walls. If you are dealing with a fully warmed up engine, then heat will be going into the intake air during the intake stroke and for the initial part of the compression stroke because the cylinder walls are at a higher temperature than the intake air (i.e. T1 gets higher than ambient), but then late in the compression stroke, heat will be going from the air into the cylinder walls. At low RPM, the heat transfer is more significant than it is at high RPM because there is more time for it to happen.
#2: The quoted compression ratio for your engine is a theoretical number based on the maximum cylinder volume (piston all the way to the bottom). But ... the intake valve doesn't close exactly at bottom-dead-center. At low revs (cranking, idle) the piston will push part of the charge backwards out of the intake valve during the first part of the compression stroke, which has the effect of reducing the "effective" compression ratio. At very high revs near redline, the pressure drop across the intake port and intake valve is significant and the cylinder won't receive a full charge to begin with, which also has the effect of reducing the "effective" compression ratio.
"What temperature is required to ignite the diesel" - this is the self-ignition temperature of the fuel, and it can vary somewhat with fuel properties. It is not unreasonable to assume that it is somewhere near 500 C, though.
