Prove that $\displaystyle |a \ \times \ b |^2 = det \left(\begin{array}{cc}(a \cdot a)&(a \cdot b)\\(a \cdot b)&(b \cdot b)\end{array}\right) $

So far I have $\displaystyle det \left(\begin{array}{cc}(a \cdot a)&(a \cdot b)\\(a \cdot b)&(b \cdot b)\end{array}\right) = (a \cdot a) \cdot (b \cdot b) - (a \cdot b) \cdot (a \cdot b)$

But I don't know how the cross product would look like after calculation. Hints?