why k times differential of functional $u:U\to \mathbb{R}$ lies in $R^{n^k}$ Let $u$ be a smooth functional $u:U\to \mathbb{R}$ ,with $U\subset \mathbb{R}^n$.
Why k times differential $D^k u$ can be identified as a vector in $\mathbb{R}^{n^k}$.
For $Du$ I know how to do it that is $Du_x :T_xU\to \mathbb{R}$  hence $Du_x \in \text{Hom}(T_xU,\mathbb{R}) \cong \mathbb{R}^n$.
I don't know how to show the result for higher order?
 A: The situation is made apparent by looking at the second total derivative (the Hessian). To evaluate $D(Du_x)$ you need to specify a point $x_1\in T_xU\cong\mathbb{R}^n$ and a tangent vector $x_2\in \mathbb{R}^n$. Indeed, the Hessian is defined as $H_x(x_1,x_2) = D(Du_x)_{x_1}(x_2)$, which is easily seen to be a bilinear map $H_x:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$. Taking another total derivative defines a trilinear ($3$-multilinear) map, and in general $D^ku: U\to L^k(\mathbb{R}^n\times\dotsm\times\mathbb{R}^n,\mathbb{R})$ associates to each $x\in U$ a $k$-multilinear map defined by the formula
$$(x_1,\dotsc, x_k)\mapsto D(D(\dotsm(Du_x)_{x_1})\dotsm)_{x_{k-2}})_{x_{k-1}}(x_k).$$
Now the problem reduces to proving that the dimension of $L^k(\mathbb{R}^n\times\dotsm\times\mathbb{R}^n,\mathbb{R})$ over the field $\mathbb{R}$ is $n^k$. To do this, fix a basis $e_1,\dotsc, e_n$ of $\mathbb{R}^n$ . Then the set of $A_{j_1\dotsc j_k}:(\mathbb{R}^n)^k\to \mathbb{R}$ defined by $A_{j_1\dotsc j_k}(e_{i_1},\dotsc, e_{i_k}) = \delta_{i_1j_1}\dotsm\delta_{i_kj_k}$ for $i_r,j_s\in \{1,\dotsc,n\}$ and $r,s\in\{1,\dotsc, k\}$ forms a basis for the set of $k$-multilinear maps; for each $j_s$ you choose from among $n$ options, and we do this $k$ times ($s=1,\dotsc, k$), so the dimension is $n^k$ as claimed.
A: Let us consider the more general case of a smooth function $f : U \to \mathbb R^m$. Then $D_x f : T_x U \to T_{f(x)} \mathbb R^m$.
To get $D^2f$ it is not sufficient to separately consider the linear maps $D_x f: T_xU \to T_{f(x)} \mathbb R^m = \mathbb R^m$, $x \in U$, but the function $D f : TU \to \mathbb R^m, Tu(x,v) = D_xf(v))$, on the tangent bundle of $U$. We have $TU = U \times \mathbb R^n$, thus
$$D f : U \times \mathbb R^n \to \mathbb R^m .$$
This can be identfied with
$$\hat D f : U \to \text{Hom}(\mathbb R^n,\mathbb R ^m) \cong \mathbb{R}^{n \cdot m} , \hat D f (x) (v) = D_xf(v) .$$
Now you see that $D^2_x f : T_x U \to T_{f(x)} \mathbb{R}^{n \cdot m}  = \mathbb{R}^{n \cdot m}$. Iterating this yields
$$D^k_x f : T_x U \to \mathbb{R}^{n^k \cdot m} .$$
