Prove that there is a natural isomorphism between $V$ and $(V^*)^*$

The isomorphism you are looking for is given by $\Phi:V\to V^{**}$ by $v\mapsto(\lambda\mapsto\lambda(v))$, i.e. you associate to an element $v$ of $V$ the element of the dual of the dual mapping the element $\lambda$ of $V^*$ (a mapping $\lambda:V\to k$) to $\lambda(v)$.

I know it can look very confusing at first, but you'll get used to it after reading it a couple of times and using it in various exercises and proofs.

For every vector $v \in V$, you have to produce some $i(v) \in V^{**}$. And this one must be a linear map

$$ i(v) : V^* \longrightarrow \mathbb{K} \ . $$

So, you have to produce an element of $\mathbb{K}$, for every $\omega \in V^*$. That is

$$ \omega \mapsto i(v)(\omega ) \ . $$

Now, having as all data a linear form $\omega \in V^*$ and a vector $v\in V$, how would you obtain, in the more natural way, an element of $\mathbb{K}$?