Vector space over complex numbers is also a vector space over the real numbers So I know I need to prove the kajillion axioms of vector space like commutativity, associativity, the additive/multiplicative identities/inverses etc. How would I go about getting started?
 A: Let $V$ be a vector space over $\mathbb{C}$.  Then for any subfield $K \subset \mathbb{C}$, $V$ is also a vector space over $K$.  Proof:
All the relevant axioms of a vector space definition are

Distributivity of scalar multiplication with respect to vector addition : $a(u + v) = au +av$
  Distributivity of scalar multiplication with respect to field addition : $(a + b)v = av + bv$
  Compatibility of scalar multiplication with field multiplication : $a(bv) = (ab)v$

(for all $a,b \in K, v\in V$)
The first one is satisified since it's already satisfied for all $a\in \mathbb{C}$ and $K \subset \mathbb{C}$.  Actually they're all satisfied because $a,b \in K \subset \mathbb{C}$.  Then since $K$ is a field by definition, we're done.
A: At your level, I suggest you prove $\mathbb{C}$ is also a vector space (of dimension 2) over $\mathbb{R}$ first. Once you "go through" the axioms and see how easy the proof is, then apply the same argument to any complex vector space. This is more complicated than it should be, but might be more helpful to you. 
