Why is the intersection number $h^2$ equal to the degree of a vector bundle of rank 2?

Question

I am trying to understand theory about ruled surfaces and am reviewing Proposition 3.18 from Arnaud Beauville's book "Complex Algebraic Surfaces":

Proposition III.18. Let $S=\mathbb{P}_{C}(E)$ be a geometrically ruled surface over $C$, $p:S\longrightarrow C$ the structure map. Write $h$ for the class of the sheaf $\mathcal{O}_S(1)$ in $Pic(S)$ (or in $H^{2}(S,\mathbb{Z})$). Then

1. $Pic(S)=p^{*}(Pic(C))\oplus\mathbb{Z}h$.
2. $H^{2}(S,\mathbb{Z})=\mathbb{Z}f\oplus\mathbb{Z}h$, where $f$ is the class of a fibre.
3. $h^2=deg(E)$.
4. $[K]=-2h+(deg(E)+2g(C)-2)f$ in $H^{2}(S,\mathbb{Z})$.

My question is in the proof of statement 3: It has the exact sequence

$$0\longrightarrow N \longrightarrow p^{*}(E)\longrightarrow \mathcal{O}_{S}(1)\longrightarrow 0$$

and from a previous process obtains that $h.[N]=0$. From this exact sequence we obtain an isomorphism $N\otimes\mathcal{O}_{S}(1)=p^{*}(\stackrel{2}\wedge E)$, whence $[N]=-h+p^{*}(e)$, writing $e$ for the class of $\stackrel{2}\wedge E$ in $Pic(C)$. It follows that

$$h^{2}=[N].h+h.p^{*}(e)=h.p^{*}(e)=deg(E)$$

Could someone help me understand why the last equality $h.p^{*}(e)=deg(E)$?