Universal property of the tensor product Assume $\Phi:V_1^*\times ...\times V_k^* \rightarrow L(V_1,...,V_k;\mathbb{R})$ is a multilinear map. $$\Phi (w^1,...,w^k)(v_1,...,v_k)=w^1(v_1)...w^k(v_k)$$
By the universal property of the tensor product this descends to a linear map $\tilde{\Phi}: V_1^*\otimes ... \otimes V_k^* \rightarrow L(V_1,...,V_k;\mathbb{R})$ $$\tilde{\Phi}(w^1 \otimes ... \otimes w^k)(v_1,...,v_k)=w^1(v_1)...w^k(v_k)$$
And now I hate myself for asking, but how is it we conclude that $\tilde{\Phi}$ is an isomorphism, s.t. $V_1^*\otimes ... \otimes V_k^* \overset{\sim}{=} L(V_1,...,V_k;\mathbb{R})$
 A: It's not surjective in general.  Let's take $k=2$ to simplify the notation.  By the definition of $\tilde\Phi$,
$$ \tilde\Phi\Big(\sum_{i=1}^n w_i^1\otimes w_i^2\Big)(v_1,\cdot)
= \sum_{i=1}^n w_i^1(v_1)w_i^2 \in V_2^\ast $$
As $v_1$ varies in $V_1$ (but the $w_i^j$ are fixed), you just get different linear combinations of $w_1^2,\dotsc,w_n^2$.  Thus
\begin{align*}
V_1 &\to V_2^\ast \\
v_1 &\mapsto \tilde\Phi\Big(\sum_{i=1}^n w_i^1\otimes w_i^2\Big)(v_1,\cdot)
\end{align*}
has finite rank.  Since every element of $V_1^\ast\otimes V_2^\ast$ can be written as a sum of finitely many elementary tensors, this shows that the range of
$$ \tilde\Phi\colon V_1^\ast\otimes V_2^\ast \to L(V_1,L(V_2,\mathbb R)) \cong L(V_1,V_2;\mathbb R) $$
is (a subspace of) $F(V_1,L(V_2,\mathbb R))$, where $F(X,Y)$ denotes the space of finite-rank linear maps from $X$ to $Y$.  If $V_1$ and $V_2$ are both infinite-dimensional, there are more maps in $L(V_1,L(V_2,\mathbb R))$ than that.
It is true that $\tilde\Phi$ is injective.  I wrote up some short notes on tensor products of vector spaces (pdf) a few years back that include this: see §40 for the statement about dual spaces; the relevant proof, for a more general statement, is in §30.  It uses the idea of hoisting linear independence from the vector spaces up to the tensor product.
