Why do electrons fall from a high excitation to a lower one? If when you shine a photon into an atom for example, and this excites an electron to a higher energy level, do the electron(s) keep going higher the more light you shine, and is there an energy limit, if so why?
Secondly, if you then stop shining light, why will the electrons fall back to a lower level? Will they at all? And why? it seems arbitrary that they will unless acted on by something else.
If they do fall back, how long does it take till they do?
I am taking QM but have not reached this understanding yet.
 A: Classical electromagnetic theory could not explain the spectral lines seen in light spectra. It predicted that an electron caught around a proton would radiate continuously (due to the radial acceleration)  and fall on the proton by emitting a continuous spectrum of light. No stable hydrogen atoms would exist.
Here is the hydrogen atom:

The fixed orbits explained the observed spectra, the Balmer and Lyman series and this was one of the main pillars for inventing quantum mechanics.

copied 
Notice  that the lines have a width

If when you shine a photon into an atom for example, and this excites an electron to a higher energy level,

Let us take the hydrogen atom as an example. 
Yes, the atom will be excited if the photon has the frequency/energy of the difference in the energy levels

do the electron(s) keep going higher the more light you shine, and is there an energy limit, if so why?

It is not a matter of quantity but of the appropriate energy differences 
Look at the energy levels . For a photon to hit an electron and ionize the atom it has to have a frequency h*nu=13.6eV. To transit to an intermediate level, the photon has to have the difference in energies, otherwise it just scatters off the field of the atom and leaves it intact.
If you give a photon the correct steps in energy , then the electron can step up until ionization. Again, it is the energy or the photon that has to match the energy level differences.
If you have a hydrogen gas, and a source with the appropriate energy levels to excite the atom then the more photons, the more excited states. General frequencies will ony by chance fit the correct difference.

Secondly, if you then stop shining light, why will the electrons fall back to a lower level? 

There is a calculable quantum mechanical probability for the electrons to fall to an empty lower level, because it is a law of nature, quantum mechanical and classical, that everything goes to the lowest allowed energy level  where stability exists. 

Will they at all? And why? it seems arbitrary that they will unless acted on by something else.

The electron  can either cascade down the levels releasing photons with the appropriate frequency/energy, or go in one step to the lowest energy level. All these are calculable probabilities in the quantum mechanical frame. Your "acted on by something else" translates in saying that the Schrodinger solutions are not enough to give widths to the predicted lines, which is true, one needs quantum electrodynamics. 
A single atom in space, if in an excited state has a calculable probability to fall to the ground state ,which is modeled by  using interactions with the QED vacuum.

If they do fall back, how long does it take till they do?

When you study further you will see that this is connected with the width of the line, which is not a strict energy line  but has a Δ(E) again calculable. These calculations need quantum field theory , not the simple Schrodinger model, as valerio92 quotes in his answer. The Heisenberg uncertainty principle then ties the decay times with this energy width.
A: In this article you can find the reason for spontaneous emission.
The excitation of an electron depends of course on the frequency of the photons you shine on the atom. In practice, these photons are wave packets. If we assume the outer electron is excited the photons must have energies that are equal to the difference (of which there are many) of the energy levels of the electron. If we take one frequency (or better said, a wave packet around a mean value for the frequency), associated with the energy difference between the first energy level of the electron and the second (below that energy the photon can't excite the electron), the more photons you shine on the atom, the higher the energy state of the electron (you have to shine them of course at a faster rate than the rate at which they fall back to lower energy levels). So by shining a lot of photons with the just amount of energy, the electron will eventually be knocked off the atom (the spreading of energies becomes less the higher the energy states).
A: The answer is thermodynamics, and the assumption that you're working in a colder environment than the temperature corresponding to a Planck distribution where your photons would be "on average" fairly present.  In other words, inside a star, where it is hotter, the atoms are NOT in their ground state most of the time - in fact, if it is hot enough, they are in their "highest state" which is an ionised state: you have a plasma.  It is simply because most atomic matter has energy levels with differences that are much larger than the average photon energies at "room temperature" (about 26 milli-eV) that we tend to say that atoms and molecules are in their ground states.  It is because at these low temperatures, it is statistically favorable to have energy spread out more than in concentrated excited states.
BTW, you can see that with rotational states of molecules: at room temperature, these are usually NOT in their ground state and excited rotational states don't "decay to ground state".  It is because their energy levels are below 26 meV.
So when you "shine light on an atom" in a cold environment, you put it out of thermodynamic equilibrium, and it will tend back to equilibrium which is its ground state.  When you "shine light on an atom" in a hot environment, it will not fall back to its ground state, because that's not its equilibrium state.
An atom in a cold environment will decay to ground state through spontaneous emission, which has an exponential time decay that is depending on the specific state and is quite difficult to calculate.
A: I want to add about spontaneous emission. Excited states of atoms are not stationary states because of atoms are not isolated QM systems. There always is interaction with electromagnetic field. Schrodinger equation for atoms in simplest form takes into account only Coulomb interaction between electrons and nucleus. In this simplified approach excited states are stationary and spontaneous emission has no place.
A: *

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do the electron(s) keep going higher the more light you shine (...)?

No, because energy levels are quantized.
This means that no matter how many photons you throw at the electron (i.e. the intensity of the light source), it won't jump to an higher energy level unless the frequency ($\nu$) of the photons is right, i.e. if
$$
\nu = \frac{\Delta E} h
$$
where $\Delta E$ is the energy difference between the energy levels and $h$ is Planck's constant.


*

*
Secondly, if you then stop shining light, why will the electrons fall
  back to a lower level? Will they at all? And why? it seems arbitrary
  that they will unless acted on by something else.

Yes, they do fall back, and the reasons are two:


*

*An atom is never truly isolated, and it will interact with the external electromagnetic field.

*Even if we assume that the atom is in free space, far from any source of EM field, it will still be subject to vacuum fluctuations of the EM field and thus eventually decay to a lower energy level. This process, which is called spontaneous emission, cannot be explained if the EM field is treated as a classical object, and its description requires the formalism of quantum field theory. For a more detailed discussion, see for example the Wikipedia page:



Spontaneous transitions were not explainable within the framework of the Schroedinger equation, in which the electronic energy levels were quantized, but the electromagnetic field was not. Given that the eigenstates of an atom are properly diagonalized, the overlap of the wavefunctions between the excited state and the ground state of the atom is zero. Thus, in the absence of a quantized electromagnetic field, the excited state atom cannot decay to the ground state. In order to explain spontaneous transitions, quantum mechanics must be extended to a quantum field theory, wherein the electromagnetic field is quantized at every point in space. The quantum field theory of electrons and electromagnetic fields is known as quantum electrodynamics.
In quantum electrodynamics (or QED), the electromagnetic field has a
  ground state, the QED vacuum, which can mix with the excited
  stationary states of the atom. As a result of this interaction, the
  "stationary state" of the atom is no longer a true eigenstate of the
  combined system of the atom plus electromagnetic field. In particular,
  the electron transition from the excited state to the electronic
  ground state mixes with the transition of the electromagnetic field
  from the ground state to an excited state, a field state with one
  photon in it. Spontaneous emission in free space depends upon vacuum
  fluctuations to get started.



*

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If they do fall back, how long does it take till they do?

The probability that the transition has not happened at time $t$ is $1-p$, where $p$ is the probability that it has happened. To calculate the transition probability per unit time, you can use the Einstein coefficients.
