The title pretty much says it all -- I'm wondering if the set of functions from $\mathbb{R}^n$ to itself, $^{\mathbb{R}^n}\mathbb{R}^n$, is isomorphic to $(^{\mathbb{R}^n}\mathbb{R})^n$, the cartesian product of $^{\mathbb{R}^n}\mathbb{R}$ with itself $n$ times.
The motivating example here is in vector calculus, where (for example) we may view the function $\mathcal{V}:\mathbb{R}^2\rightarrow\mathbb{R}^2$ defined by $$\mathcal{V}(x,y)=\begin{bmatrix} x^2y^2 \\ e^xe^y \end{bmatrix}$$ as two functions $f:\mathbb{R}^2\rightarrow\mathbb{R}$ and $g:\mathbb{R}^2\rightarrow\mathbb{R}$ with $f(x,y)=x^2y^2$ and $g(x,y)=e^xe^y$. It seems to me that we could very naturally define operations like $\nabla\circ$,$\nabla\times$ and so on with relative ease in $(^{\mathbb{R}^2}\mathbb{R})^2$ as coordinate wise operations. Is this a well established relationship, and if so are there any texts that exploit this relationship to treat vector calculus like 'iterated multivariate calculus'?
More generally, $$^X(Y^n)\cong\left(^XY\right)^n$$
Define $[n]=\{1,2,\dots,n\}$. Then: $Y^n\cong {}^{[n]}Y$ and the result follows from the general rule:
$$^X\left({}^YZ\right)\cong {}^{X\times Y}Z\equiv {}^Y\left({}^XZ\right)$$