I've been trying to show that if $V_1,\dots,V_k,W$ are vector spaces over $K$, then
$$\mathcal{L}(V_1,\dots,V_k;W)\simeq \mathcal{L}(V_1;\mathcal{L}(V_2,\dots,V_k;W)),$$
I think I've got the idea, but I've got confused with notation, so I'm unsure if it's right. The idea is: pick $f\in \mathcal{L}(V_1,\dots,V_k;W)$, if I fix the first argument, I have a function from in $\mathcal{L}(V_2,\dots,V_k;W)$. So that the map that picks $v_1\in V_1$ and gives the function $f$ with $v_1$ fixed is in $\mathcal{L}(V_1;\mathcal{L}(V_2,\dots,V_k;W))$, so the maps that picks $f$ and gives this function should be the isomorphism.
So I do as follows: let $f \in \mathcal{L}(V_1,\dots,V_k;W)$, define $T_f \in \mathcal{L}(V_1;\mathcal{L}(V_2,\dots,V_k;W))$ be given by $T_f(v_1)=f(v_1,\cdot,\dots,\cdot)$. Define then $\psi : \mathcal{L}(V_1,\dots,V_k;W)\to\mathcal{L}(V_1;\mathcal{L}(V_2,\dots,V_k;W))$ by the rule $\psi(f)=T_f$, then $\psi$ is isomorphism.
To show that, we notice that $\psi$ is linear, for $\psi(\lambda f + g)=T_{\lambda f+g}$, and then we have that $T_{\lambda f+g}(v_1)=(\lambda f+g)(v_1,\cdot,\dots,\cdot)$, so that by definition of operations between multilinear maps we have $T_{\lambda f+g} = \lambda T_f + T_g$ and then $\psi(\lambda f + g)=\lambda \psi(f) + \psi (g)$.
Injectivity is shown by noticing that $\psi(f)=0$ implies that $T_f=0$, so that we have $T_f(v_1)=0$ for all $v_1\in V_1$. This means that $f(v_1,\cdot,\dots,\cdot)=0$ and so this implies $f=0$, hence $\psi$ is injective.
Surjectivity we show by the following process, let $\varphi \in \mathcal{L}(V_1;\mathcal{L}(V_2,\dots,V_k;W))$ then $\varphi(v_1)$ is a multilinear mapping from $V_2,\dots,V_k$ into $W$. Define $\eta(\varphi)(v_1,\dots,v_k) = \varphi(v_1)(v_2,\dots,v_k)$, then $\eta(\varphi)$ is multilinear from $V_1,\dots,V_k$ into $W$, more than that, we have
$$\psi(\eta(\varphi))=T_{\eta(\varphi)}\Longrightarrow T_{\eta(\varphi)}(v_1)=\eta(\varphi)(v_1,\cdot,\dots,\cdot)=\varphi(v_1)(\cdot,\dots,\cdot),$$
so that $\psi$ is surjective and hence bijective. Since $\psi$ is linear, injective and surjective, $\psi$ is isomorphism.
Now, is this proof correct? I'm a little unsure mainly because of notation, and in the surjectivity part.
Thanks very much in advance!
You're asking to show $\operatorname{Mult}(V_1 \times \ldots \times V_n,W) = \operatorname{Hom}(V_1,\operatorname{Mult}(V_2\times \ldots \times V_n,W))$? We have
$$\begin{eqnarray*} \operatorname{Mult}(V_1 \times \ldots \times V_n,W) &=& \operatorname{Hom} ( V_1 \otimes \ldots \otimes V_n,W) \\ &=&\operatorname{Hom}(V_1,\operatorname{Hom}(V_2 \otimes \ldots \otimes V_n,W))\\ &=&\operatorname{Hom}(V_1,\operatorname{Mult}(V_2\times \ldots \times V_n,W))\end{eqnarray*}$$
showing your result. In the first line we used the universal property of the tensor product, second line tensor-hom adjunction, third line the universal property again.