Why can $D^ku(x) := \{D^\alpha u(x) \mid |\alpha| = k \}$ be regarded as a point in $\mathbb{R}^{n^k}$? This is a comment made in the Appendix of Evans's Partial Differential Equations. He defines the set of all $k$ order partial derivatives as $D^ku(x):= \{D^\alpha u(x) \mid |\alpha| = k \}$ (using multi-index notation) and considers them as points in $\mathbb{R}^{n^k}$ after "assigning some ordering to the various partial derivatives."
As a simple case, $k=2$, the set of all partial derivatives $D^2u$ is given, for example, by the Hessian. I notice that there are $n^2$ possible such partial derivatives, but I'm not exactly sure what the relationship is between $D^ku$ and points in $\mathbb{R}^{n^k}$.
 A: This is just notational convenience that extends the basic idea for the gradient.  For example, if $u: \Omega \to \mathbb{R}$ is a differentiable function on $\Omega \subset \mathbb{R}^n$ then we have $\nabla u : \Omega \to \mathbb{R}^n$ and for each $x \in \Omega$ we view $\nabla u(x) \in \mathbb{R}^n$ as a point in the space $\mathbb{R}^n$. 
Now, for the second derivatives we can view the Hessian matrix $D^2 u: \Omega \to \mathbb{R}^{n \times n}$ at each point as a matrix $D^2 u(x) \in \mathbb{R}^{n \times n}$ or we can identify $\mathbb{R}^{n \times n} \simeq \mathbb{R}^{n^2}$ and view $D^2 u(x) \in \mathbb{R}^{n^2}$.  
For the third derivatives we have $n^3$ terms, so $D^3 u : \Omega \to \mathbb{R}^{n^3}$, for the fourth derivative we have $D^4 u : \Omega \to \mathbb{R}^{n^4}$, and so on.  This is all that is being done in this notation: $D^k u : \Omega \to \mathbb{R}^{n^k}$.
The reason for the "after some ordering" comment is that there is no canonical choice of how to map $\mathbb{R}^{n\times n}$ to $\mathbb{R}^{n^2}$, nor is there one particular way for enumerating all of the third, fourth, or kth order derivatives.  So, we just choose some consistent way of ordering them in order to view them as above.
