Proof of the generalized Clairaut's theorem. I'm trying to prove the following preposition:
Clairaut's basic theorem says that if $f: \mathbb{R^n} \to \mathbb{R}$ is a $C^2$ function, then $\partial_{ij}f = \partial_{ji}f$ for all $1 \leq i, j, \leq n$.
Use the basic theorem to prove the more generalized version of Clairaut's theorem of Clairaut's theorem. That is, if $f: \mathbb{R^n} \to \mathbb{R}$ is a $C^k$ function, then $\partial_{i_1, ..., i_k} = \partial_{j_1, ..., j_k}f$ whenever $(i_1, ..., i_k)$ and $(j_1, ..., j_k)$ are tuples of indices which are re-arrangements of each other.
I'm attempting to prove this by induction. I could use some help in proving this. Is Induction the best way to go for this? Perhaps there's a better way?
Proof by Induction.
Base case: Let $n = 1$, so that $i = j = 1$. 
Since $f$ is of type $C^k$, it follows that $C^k \subseteq C^2$.
Then by the basic Clairaut's theorem,  $\partial_{ij}f = \partial_{ji}f$ holds.
Suppose for all $1 < i,j \leq m$ that $\partial_{i_1, ..., i_m} = \partial_{j_1, ..., j_m}$, for $\{i_1, ..., i_m\}$ and $\{j_1, ..., j_m\}$ being re-orderings of each other.
Consider $m + 1$, then $\partial_{i_1, ..., i_m, i_{m+1}} = \partial_{j_1, ..., j_m, j_{m+1}}$
I am trying to consider transpositions of the indices in such a way that allows me to use my inductive hypothesis. If I swap the last $i_{m+1}$ with $i_1$, but I am not sure how to invoke the IH carefully. By swapping the first and the last i's, I can show use the IH on the first m j terms, but then I am stuck trying to show it for the m+1 j'th term.
Any help appreciated.
 A: This is not a problem about differentiation, but a problem in discrete mathematics.
Denote by $W_r$ the set of words $\alpha=\alpha_1\alpha_2\ldots\alpha_r$ of length $r\geq0$ over the alphabet $[n]$. For an $\alpha\in W_r$ denote by $\hat\alpha$ the associated multiset on $[n]$, i.e., the function $\hat\alpha:\>[n]\to{\mathbb N}_{\geq0}$ giving the multiplicity with which each letter (coordinate number) $j$ occurs in $\alpha$:
$$\hat\alpha(j):=\#\bigl\{ k\in[r]\>\bigm|\>\alpha_k=j\bigr\}\qquad(1\leq j\leq n)\ .$$
Two words $\alpha$, $\beta\in W_r$ with $\hat\alpha=\hat\beta$ are called equivalent.
For $\alpha\in W_r$ denote by $D^\alpha f$ the $r^{\rm th}$ partial derivative of $f$ with respect to $x_{\alpha_1}$, $x_{\alpha_2}$, $\ldots$, $x_{\alpha_r}$ in this order. We then have to prove the  propositions
$${\cal P}_r:\qquad  \alpha, \beta\in W_r\quad\wedge\quad   \alpha\sim \beta\qquad\Rightarrow\qquad D^\alpha f=D^\beta f\ .$$
Since ${\cal P}_0$ and ${\cal P}_1$ are trivially true we may assume $r\geq2$, and that ${\cal P}_s$ is true for $s<r$. Take two equivalent words $\alpha$, $\beta\in W_r$. Then $$\alpha=\alpha'\alpha_r\>, \quad\beta=\beta'\beta_r\>, \qquad \alpha', \beta'\in W_{r-1}\ .$$
If $\alpha_r=\beta_r$ then $\alpha'\sim\beta'$. The induction hypothesis then implies$$D^\alpha f={\partial\over\partial x_{\alpha_r}}D^{\alpha'}f={\partial\over\partial x_{\alpha_r}}D^{\beta'}f=D^\beta f\ .$$
If $\alpha_r\ne\beta_r$ then $\beta_r$ has to occur in $\alpha'$, and we can say the following: There is $\alpha''\in W_{r-2}$ such that $\alpha'\sim\alpha''\beta_r$, and similarly, there is $\beta''\in W_{r-2}$ such that $\beta'\sim\beta''\alpha_r$. From $\alpha\sim\beta$ it follows that $\alpha''\sim\beta''$. By the induction hypothesis we then have
$$D^\alpha f={\partial\over \partial x_{\alpha_r}}D^{\alpha'} f={\partial\over\partial x_{\alpha_r}}D^{\alpha''\beta_r} f={\partial^2\over\partial x_{\beta_r} \partial x_{\alpha_r}}D^{\alpha''} f\ ,\tag{1}$$
and similarly
$$D^\beta f={\partial^2\over\partial x_{\alpha_r} \partial x_{\beta_r}}D^{\beta''} f\ .\tag{2}$$
The induction hypothesis and $\alpha''\sim\beta''$ imply that $D^{\alpha''} f=D^{\beta''} f$. Clairaut's original version of the theorem  then allows to conclude that the right hand sides of $(1)$ and $(2)$ are equal.
