How to prove exponential correspondence for tensors? Let ${\rm Vec}_{\mathbb{R}}$ denote the category of finite dimensional vector spaces over the real numbers. Let ${\rm Hom}_k(V,W)$ denote the space of multilinear maps $\underbrace{V\times\dots\times V}_{k\ times} \to W$ and $V_k^1$ denote the space of multilinear maps $\underbrace{V\times\dots\times V}_{k\ times} \times V^* \to \mathbb{R}$.
I want to show that there is a natural isomorphism between $Hom_k(V,V)$ and $V_k^1$. I have reduced the problem to showing the existence of a natural isomorphism between $Hom_k(V,V^{**})$ and $V_k^1$. Intuitively it is clear to me that such an identification can be accomplished by currying, that this is a vector space isomorphism should be verifiable simply by checking the definitions, and that naturality should follow from (some variant of) the Yoneda Lemma. However, I am having some difficulty finding the endofunctors in $Vec_{\mathbb{R}}$ that would allow me to verify all of these claims.
Reduction: I know that, given a finite dimensional vector space $V$, there is a natural isomorphism with $V^{**}$. Using hom functors, it isn't that difficult to extend this to a natural isomorphism between $Hom_k(V,V)$ and $Hom_k(V,V^{**})$. 
What I have tried: I have tried to find functors $$f_1: Vec_{\mathbb{R}} \to Vec_{\mathbb{R}}, V \mapsto Hom_k(V,V^{**})$$ and $$f_2: Vec_{\mathbb{R}} \to Vec_{\mathbb{R}}, V\mapsto V_k^1.$$ However, given vector space homomorphisms $T:V\to W$, it is unclear to me how to find vector space homomorphisms $Hom_k(V,V^{**}) \to Hom_k(W,W^{**})$ or $Hom_k(W,W^{**}) \to Hom_k(V,V^{**})$ which would give the morphism part of the functors. First I tried using a contravariant hom functor, i.e. pre-composition by T, $$T \mapsto f(T(\cdot),\dots,T(\cdot)),$$ but the problem with this is that it only maps $Hom_k(W,W^{**}) \to Hom_k(V,W^{**})$, but clearly I need a map $Hom_k(W,W^{**}) \to Hom_k(V,V^{**})$. 
So then I tried restricting from $Vec_{\mathbb{R}}$ to the subcategory $\mathbb{V}$, defined to be the category whose objects are all $n$-dimensional vector spaces over $\mathbb{R}$ and whose morphisms consist only of invertible linear transformations, and then finding functors $$f_1: \mathbb{V} \to Vec_{\mathbb{R}}, \quad f_2: \mathbb{V} \to Vec_{\mathbb{R}},$$ with the same rules of assignment for the objects as before, namely $f_1: V \mapsto Hom(V,V^{**})$ and $f_2: V \mapsto V_k^1$. Then I tried the following rule of assignment for morphisms for $f_1$: $$T\mapsto T^{-1} f(T(\cdot),\dots,T(\cdot))$$ -- this does map $Hom_k(W,W^{**})\to Hom_k(V,V^{**})$ as desired, but it turns out this satisfies neither associativity nor functoriality. Now I am stuck. I haven't tried to find a rule of assignment for morphisms for $f_2$, but it seems like one would run into the analogous problems.
However, without two endofunctors $f_1$ and $f_2$, I can't show that there is a natural transformation between them which is also an isomorphism and which is a vector space isomorphism when applied to objects.
In terms of exponential objects, I think what I want to show is equivalent to: $$ V^{V^k}   \cong \mathbb{R}^{V^k \otimes V^*} .$$ Then what I have reduced it to is showing $$ V^{V^k}   \cong (\mathbb{R}^{V^*})^{V^k} \cong \mathbb{R}^{V^k \otimes V^*} .$$ Now the left isomorphism is essentially easy, since $V \cong V^{**}:=\mathbb{R}^{V^*}$, so what I need to show is the right isomorphism, which is the vector space version of the natural exponential object correspondence. So if I can prove that $(X^Y)^Z \cong X^{Y \times Z}$ in $Set$, then maybe the proof will carry over almost word for word for the corresponding version in $Vec_{\mathbb{R}}$. I still need to look into this line of approach more, although so far I have not succeeded yet in demonstrating the naturality of the isomorphism $(X^Y)^Z \cong X^{Y \times Z}$ in $Set$, so it might not be as easy as I am hoping.
EDIT: If I can show that $Vec_{\mathbb{R}}$ is Cartesian closed, then this should follow from Exercise 1(b), section 6.6., p.123, of Awodey's Category Theory. Although the exercise is only to prove that we have an isomorphism, $(A^B)^C \cong A^{B \times C}$, but does not include a proof of naturality.
 A: Your "functors" aren't actually functors. I'll stick to standard notations because "$\operatorname{Hom}_k(V,V)$" is too easy to mistake for the space of $k$-linears maps $V \to V$ for some field $k$... I'll also let the base field $\mathbb{R}$ be implied throughout.
Let's do the simplest case, $k=1$. Then you have a bifunctor
$$\operatorname{Hom} : \mathsf{Vec}^{\mathrm{op}} \times \mathsf{Vec} \to \mathsf{Vec}, \; (V,W) \mapsto \operatorname{Hom}(V,W)$$
It's not possible to compose that with the functor $\mathsf{Vec} \to \mathsf{Vec} \times \mathsf{Vec}$, simply because the (co)domains don't match... So the mapping $V \mapsto \operatorname{Hom}(V,V)$ doesn't yield a functor a priori. I can't see any reasonable way to make that into a functor. Your question is doomed from the start...

However you do have the bifunctor $\operatorname{Hom}$ as above. You also have another bifunctor, say $$\Phi : \mathsf{Vec}^{\mathrm{op}} \times \mathsf{Vec} \to \mathsf{Vec}, \; (V,W) \mapsto \operatorname{Hom}(V \otimes W^*, \mathbb{R}).$$
This is indeed a bifunctor; given $f : V' \to V$ and $g : W \to W'$, you get $\Phi(f,g) : \Phi(V,W) \to \Phi(V',W')$ given by $t \mapsto t \circ (f \otimes g^*)$ (where $g^* : \operatorname{Hom}(W', \mathbb{R}) \to \operatorname{Hom}(W,\mathbb{R})$ is precomposition by $g$).
These two functors are naturally isomorphic, when you restrict to finite-dimensional spaces. Indeed define a natural transformation $\eta : \operatorname{Hom} \to \Phi$ by:
\begin{align}
\eta_{(V,W)} : \operatorname{Hom}(V,W) & \to \operatorname{Hom}(V \otimes W^*, \mathbb{R}) \\
t & \mapsto (\eta(t) : v \otimes \psi \mapsto \psi(t(v)))
\end{align}
It's not hard (but it's a bit tedious) to check that this is a natural transformation. Both spaces have the same dimension, namely $(\dim V) (\dim W)$. So to check that this is an isomorphism for all $(V,W) \in \mathsf{Vec}^\mathrm{op} \times \mathsf{Vec}$, it suffices to check that this is surjective.
For a fixed couple $(V,W)$, let $(w_1, \dots, w_n)$ be a basis of $W$ (recall that it's finite dimensional), and let $(w_1^*, \dots, w_n^*)$ be the dual basis of $W^*$. Suppose given some $\beta \in \operatorname{Hom}(V \otimes W^*, \mathbb{R})$. Then you can let $\alpha \in \operatorname{Hom}(V, W)$ be defined by
$$\alpha(v) = \sum_{i=1}^n \beta(v \otimes w_i^*) \cdot w_i.$$
It is now a very tedious (but completely mechanical) check that $\eta_{(V,W)}(\alpha) = \beta$. The mapping is surjective between spaces of the same finite dimension, hence it's an isomorphism.

Finally, to get the general case, consider the functor $\bigotimes : \mathsf{Vec}^{\times k} \to \mathsf{Vec}$ given by $\bigotimes(V_1, \dots, V_k) = V_1 \otimes \dots \otimes V_k$, and compose it with the natural isomorphism $\eta$ to get a natural isomorphism
$$\operatorname{Hom}(V_1 \otimes \dots \otimes V_k, W) \xrightarrow{\cong} \operatorname{Hom}(V_1 \otimes \dots \otimes V_k \otimes W^*, \mathbb{R}).$$
But again this is contravariant in the $V_i$ and covariant in $W$, so you can compose with $V \mapsto (V, \dots, V)$ because this is covariant in everything...
A: $\newcommand{\Vec}{\mathsf{Vec}}$$\newcommand{\Hom}{\operatorname{Hom}}$$\newcommand{\id}{\operatorname{id}}$My answer exceeded the character limit, so here's the last part:
Finally, to get the general case, consider the following functor: $$\bigotimes: \Vec^{\times k} \to \Vec, $$ which acts by: $$\bigotimes(V_1, \dots, V_k) = V_1 \otimes \dots \otimes V_k. $$

Claim: $\bigotimes$ is actually a functor.

We already gave the proposed object part above ($\bigotimes(V_1, \dots, V_k) = V_1 \otimes \dots \otimes V_k$), now the morphism part is proposed to be the following: $$\bigotimes(f_1, \dots, f_k) = f_1 \otimes \dots \otimes f_k, $$ perhaps unsurprisingly. Now we need to check that this actually constitutes a functor.
Let $\id_{(V_1,\dots,V_k)}=(\id_{V_1}, \dots, \id_{V_k})$ (this holds by definition of product category). Then: $$\bigotimes \id_{(V_1, \dots, V_k)} = \bigotimes (\id_{V_1}, \dots, \id_{V_k}) = \id_{V_1} \otimes \dots \otimes \id_{V_k}. $$ For $\bigotimes$ to be a functor, it is necessary that this be the identity morphism on $\bigotimes (V_1, \dots, V_k) = V_1 \otimes \dots \otimes V_k$. However, by definition of the tensor product, it is actually relatively clear that: $$\id_{V_1} \otimes \dots \otimes \id_{V_k}=\id_{V_1 \otimes \dots \otimes V_k} = \id_{\bigotimes(V_1, \dots, V_k)} . $$ Thus $\bigotimes$ preserves identity morphisms, $$\bigotimes \id_{(V_1, \dots, V_k)} = \id_{\bigotimes (V_1, \dots, V_k)}, $$ as required. Now we need to show that $\bigotimes$ is compatible with composition of morphisms, i.e. given: $$(V_1, \dots, V_k) \overset{(f_1, \dots, f_k)}{\to} (V_1', \dots, V_k') \quad \text{and} \quad (V_1',\dots, V_k') \overset{(g_1, \dots, g_k)}{\to} (V_1'', \dots, V_k''), $$ one has that: $$\bigotimes\left( (g_1, \dots, g_k) \circ (f_1, \dots, f_k)  \right) = \bigotimes(g_1, \dots, g_k) \circ \bigotimes (f_1, \dots, f_k). $$ Evaluating the left-hand side first, we have: $$\bigotimes \left( (g_1, \dots, g_k) \circ (f_1, \dots, f_k)  \right) = \bigotimes (g_1\circ f_1, \dots, g_k \circ f_k)= (g_1 \circ f_1) \otimes \dots \otimes (g_k \circ f_k). $$ Now evaluating the right-hand side, one finds: $$\bigotimes (g_1, \dots, g_k) \circ \bigotimes (f_1, \dots, f_k) = (g_1 \otimes \dots \otimes g_k) \circ (f_1 \otimes \dots \otimes f_k). $$ We showed, when proving that $\Phi$ is a functor, that the tensor product $\otimes$ plays nicely with composition $\circ$, so we have finally that: $$(g_1 \otimes \dots \otimes g_k) \circ (f_1 \otimes \dots \otimes f_k) = (g_1 \circ f_1) \otimes \dots \otimes (g_k \circ f_k). $$ In conclusion, we have shown that: $$\bigotimes\left( (g_1, \dots, g_k) \circ (f_1, \dots, f_k)  \right) = (g_1 \circ f_1) \otimes \dots \otimes (g_k\circ f_k) = \bigotimes(g_1, \dots,g_k ) \circ \bigotimes(f_1, \dots, f_k), $$ hence $\bigotimes$ is compatible with composition of morphisms and thus really is a functor, as claimed.
Claim: $op: \Vec \to \Vec^{op}$ is a contravariant functor.
The object part is the same as the object part of the identity functor: $op: V \mapsto V$.
For the morphism part, we have the following rule (as follows from the definition of dual category): $$op: \left( V \overset{f}{\to} V'   \right) \mapsto \left(  V \overset{f^{op}}{\gets} V'  \right). $$ Note that while $f$ is always a function (a linear transformation in fact), in general $f^{op}$ is not, i.e. only a morphism but not a function (although one can define it as a function in the case that $f$ happens to be an isomorphism).
Anyway, we need to show that $op$ is compatible with identity morphisms: $op(\id_V) = \id_{op(V)}=\id_V$. One has that: $$op \left( V \overset{id_V}{\to} \id_V   \right) \mapsto \left( V \overset{\id_V}{\gets} V \right). $$ Now since $\id_V$ is an isomorphism, we have that the left and right hand sides are equal, hence $op(\id_V)= \id_V$, as required.
Now we show that $op$ is compatible with composition of morphisms in a contravariant manner. $$op(g \circ f) = \left(  V \overset{(g\circ f)^{op}}{\gets} V''  \right) = \left(  V \overset{f^{op}}{\gets} V'  \right) \circ \left(  V' \overset{g^{op}}{\gets} V''  \right) = op(f) \circ op(g),  $$ as expected and required.
Because we have a natural isomorphism between $\Hom(-,-)$ and $\Phi(-,-)$, it follows that we also have a natural isomorphism between $\Hom((\bigotimes(-,\dots, -))^{op}, -  )$ and $\Phi((\bigotimes(-,\dots, -))^{op},- )$, i.e. for any $k-$tuple of finite-dimensional vector spaces $(V_1, \dots, V_k)$, and any finite-dimensional vector space $W$, one has that: $$\Hom(V_1 \otimes \dots \otimes V_k, W) \cong \Hom(V_1 \otimes \dots \otimes V_k \otimes W^*, \mathbb{R} ), $$ the isomorphism being natural. 
Claim: $\times k: \Vec \to \Vec^{\times k}$ is a functor.
For the object part, we have simply: $$\times k (V) = \underbrace{(V, \dots, V)}_{k\text{ times}}, $$ and for the morphism part, given $V \overset{f}{\to} V'$, one has: $$\times k(f) = \underbrace{(V, \dots, V)}_{k\text{ times}} \overset{\underbrace{(f, \dots, f)}_{k \text{ times}}}{\to} \underbrace{(V', \dots, V')}_{k\text{ times}}.$$ Now one has fairly immediately that: $$\times k(\id_V) = (\id_V, \dots, \id_V) = \id_{(V,\dots, V)}. $$ Likewise, one also has fairly immediately that: $$\times k (g \circ f) = (g\circ f, \dots, g\circ f) = (g, \dots, g)\circ(f,\dots,f) = \times k(g) \circ \times k(f). $$ Thus $\times k: \Vec \to \Vec^{\times k}$ is a functor, as claimed.
Thus composing functors again, we get a natural isomorphism between $$\Hom\left( \left(\bigotimes( \times k (-)  )\right)^{op}, - \right) \quad \text{and} \quad \Phi\left( \left(\bigotimes(\times k (-))  \right)^{op},-   \right),$$ thus, as we wanted to show, for any choice of two finite-dimensional vector spaces $V$ and $W$, there is a natural isomorphism between: $$\Hom(\underbrace{V \otimes \dots \otimes V}_{k \text{ times}}, W) \quad \text{and} \quad \Hom(\underbrace{V \otimes \dots \otimes V}_{k\text{ times}} \otimes W^*, \mathbb{R}). $$
