Exterior Product

Question

I am studying exterior product of vector bundle, and need some indication. Let $E$ be an algebraic vector bundle of rank $r$ and degree $d$, Then $\Lambda^2 E$ is of rank $r'=r(r-1)/2$, but is of determinant 0, because $$\Lambda^{r'}(\Lambda^2 E)\subset \Lambda^{r(r-1)}E=0$$ which I can't understand! because its determinate has to be a line bundle? Is there any mistake in that calculus? Thanks