Why is the nullity of an invertible matrix 0? I am aware that nullity is the dimension of the null space. So I assume that the null space is empty? But why?
The unique vector space of dimension zero is the trivial space $\{0\}$ which includes the zero vector. This space is nonempty, and in fact the empty set is not a vector space because any vector space must have the zero vector.

By the invertible matrix theorem, one of the equivalent conditions to a matrix being invertible is that its kernel is trivial, i.e. its nullity is zero. I will prove one direction of this equivalence and leave the other direction for you to prove.

$(\Rightarrow)$ Suppose $A$ is an invertible $n\times n$ matrix. Let $v\in\ker A$ so that $Av=0$. Then $A^{-1}Av=A^{-1}0\iff Iv=0\iff v=0$. Thus, $\ker A=\{0\}$ so $A$ has nullity zero.
$(\Longleftarrow)$ Suppose that the nullity of a square matrix $(n\times n)$ A is equal to 0.

So this means that for any vector $(n\times 1)$, $Av=0$ has only $1$ solution, which is the trivial one: $v=0$.

And we know that this is equivalent to say that the determinant of A is nonzero.

And finally, because the determinant of A is nonzero, it means it is invertible

This completes the proof of equivalence as Dave already posted the $(\Longrightarrow)$.