Let $V$ be a finite dimensional real vector space, let $\mathrm{Alt}^k(V)$ denote the space of alternating $k$-linear forms on $V$ and let $\wedge^k(V)$ denote the $k^{th}$ exterior power of $V$. I am trying to see why the algebraic dual $\wedge^k(V)^* := (\wedge^k(V))^*$ is isomorphic to $\mathrm{Alt}^k(V)$. Here are my thoughts:
By the universal property of the exterior power, for any alternating $k$-linear form $f$ with domain $V^k$ there exists a unique linear form $\phi$ with domain $\wedge^k(V)$ such that $ \phi(v_1 \wedge \cdots \wedge v_k) = f(v_1, \dots, v_k). $ The universal property thus provides a mechanism to produce elements in $\wedge^k(V)^*$ from elements in $\mathrm{Alt}^k(V)$ and this mechanism of production is unique.
Thus, we have an injection $ \Phi: \mathrm{Alt}^k(V) \longrightarrow \wedge^k(V)^* $
What I'm not sure about is how to argue surjectivity; what is the best way to approach this?