This question occurred to me while thinking on another one here, [Name for an operation on matrices?](https://mathoverflow.net/q/219582/41291)


Can one define *in an invariant way* a binary operation on finite-dimensional vector spaces - let us denote it somehow suggestively by $(V,W)\mapsto V^{\otimes W}$ - with the property$$\dim(V^{\otimes W})=\dim(V)^{\dim(W)}?$$


Added later (and I should do it from the beginning as it is important): the construction from the linked question suggests that this operation seemingly should act on linear operators by assigning to $f:V\to X$ and $g:Y\to W$ certain operator (explicitly given there in terms of chosen bases)
$$
f\dagger g:V^{\otimes W}\otimes Y\to X\otimes W,
$$
with $\operatorname{rank}(f\dagger g)\geqslant\operatorname{rank}(f)\operatorname{rank}(g)$.


I'm aware that this looks even more impossible, but anyway - with bases it is done in that question.



Exponentiation of vector spaces?
Okay, so you can get pretty close as follows: I still don't think $V^{\otimes W}$ makes sense, but riffing off of your comment, we can make sense of $(1 \oplus V)^{\otimes W}$ (where $1$ denotes the $1$-dimensional vector space). The guiding intuition is the binomial expansion


$$(1 + V)^W = 1 + {W \choose 1} V + {W \choose 2} V^2 + \dots $$


which can be upgraded to the functor


$$(1 \oplus V)^{\otimes W} = 1 \oplus (W \otimes V) \oplus (\Lambda^2 W \otimes V^{\otimes 2}) \oplus \dots $$


When $V = 1$ we reproduce the exterior algebra functor. Note that there's no hope to upgrade the $GL(V)$ action on this to a $GL(1 \oplus V)$ action: each component of this direct sum contains as a factor a different irrep of $GL(W)$, so any such upgrade must continue to respect this direct sum decomposition, but this is impossible already for the summand $W \otimes V$. 


