Why is the vector of angular momentum perpendicular to the plane of rotation? I just don't understand this. I have read about the right hand rule, but I am finding it hard to take it on faith, and would like to have some proof. 
Also, try not to make your explanation too mathematical. I would prefer something intuitive and conceptual. 
 A: As with all mathematical conventions in physics, the reason why we represent something a certain way is because it is useful.
Angular velocity is a pseudovector, so its direction is defined by the axis about which an object rotates, which makes the angular velocity vector normal to the plane of rotation. Angular momentum is defined by $$\vec L = I\vec\omega$$ while this may make it appear the angular momentum is in the same direction as angular velocity, this is actually only the case when the axis of rotation is one the the principal axes of rotation, due the fact that the moment of inertia is actually represented by a matrix.
There is no need to prove this is the angular momentum, as this is how it defined. The reason why we define it this way is, as stated above, because it is useful. For example, it follows a conservation law and adds nicely.
A: We say that angular momentum is perpendicular to the plane of rotation (in a simple case like a rotating disk) only because Josiah Willard Gibbs and Oliver Heaviside popularized vector algebra (including a vector product using the right-hand rule) and vector calculus in the early 1900’s.
There are more modern formalisms — not generally taught in high school and often not in universities either — in which angular momentum is represented as a tensor which one can visualize as lying in the plane of rotation. (Or a bivector, or a two-form, or an element in Clifford algebra, etc. There are lots of formalisms for representing directional physical quantities in multidimensional spaces!)
The different formalisms describe the same physics using different mathematical objects. There is no sense in which angular momentum is “really” perpendicular to the plane of rotation. In fact, in higher dimensions Gibbs’ vector product doesn’t even make sense because there is not a unique direction perpendicular to a plane.
All of these other formalisms are more in tune with your intuition that angular momentum should be a planar quantity. They are all capable of distinguishing the “$x$ toward $y$ plane” from the “$y$ toward $x$” plane, which is necessary to keep track of which way something is rotating.
As an example, you may be interested to read about exterior algebra.
A: 1) If the body is moving in a  plane then it's position and velocity/momentum vectors are in that plane too. This is easy to verify. 
2)Angular momentum is a cross product of position and momentum
3)A cross product is always perpendicular to the plane in which the vectors to be crossed are.
4) We conclude that the angular momentum is perpendicular to the plane in which the object is moving( of which rotation is a special case).
A: The set of points unaffected by rotation forms a line, the axis of rotation, which is perpendicular to the plane of rotation. One can take this axis as defining the direction of angular momentum. However, AM, is a pseudovector . It is in reality an antisymmetric tensor and such objects have 3 components in 3D. Coincidentally this can be treated as something akin to a vector. As I said , it is really an antisymmetric tensor, $\vec r \times \vec p$, and this defines, guess what, a surface parallel to the plane of rotation, as in Kepler's second law. Does this help you to get an intuition?
