Proving euler class of trivial bundle is zero

Question

I am trying to prove that if $p:E \to B$ is a rank $n$ trivial bundle then the euler class is zero. The definition I am using for euler class is that it is a pull back of the Thom class in $H^n(E,E_0)$ to $H^n(B)$. I have looked at the explanation here but I dont really understand the equivalence of the two different descriptions he gives for the euler class. Also I have another argument but I am not sure if it is right. Since $H^n(E)$ is isomorphic to $H^n(B)$ we need to show that the class in $H^n(E)$ is zero. But we have the commutative diagram $$\begin{array}& &H^n(E,E_0)&\to &H^n(E)\\ &\uparrow & &\uparrow \\ &H^n(\mathbb R^n,\mathbb R^n-\{0\}) &\to &H^n(\mathbb R^n)\end{array}$$ But $H^n(\mathbb R^n)$ is zero and therefore the image in $H^n(E)$ is zero. The vertical arrows in the commutative diagram come from the projection $q:E \to \mathbb R^n$