Why is $\mathbb{E}[XX] = \mathbb{E}[X^2]$? Why is $\mathbb{E}[XX] = \mathbb{E}[X^2]$?
This is always presented as obvious and without explanation, but it is not obvious to me. How does one apply the principles of a multivariate distribution involving a random variable with itself?

I think you might be confused by the fact that one expression has two $X$'s while the other on has one $X$. You must realize that all of the $X$'s are the same thing. For example, let $X$ be $0$ or $1$ according to the flip of a coin, for tails and heads, respectively. If you flip the coin and get heads, then $X=1.$ This means that $X\cdot X=1\cdot1$, $X^2=1^2$, $X^5+3X^2-2X+8=1^5+3\cdot1^2-2\cdot1+8$, etc. You can't let $X$ be different things in different places. 
If you want two distinct coin flips, you have to "have two copies of $X$", call them $X_1$ and $X_2$. Now $E[X_1X_2]\neq E[X_1^2].$
Now $X_1$ and $X_2$ have a joint distribution, but $X$ does not have a joint distribution with itself. However, a function of $X$, e.g. $g(X)=X^2$ will have a different distribution that $X$ usually. But $XX$ has the same distribution as $X^2.$
