Proof That the Second Differential of a Zero Form or Function is Zero

Theorem
If 
\[\omega^0\]
  is a twice differentiable zero form in  
\[\mathbb{R}^n\]
  ie a function, then  
\[d(d \omega^0) =0\]
.
Proof
We can write  
\[\omega_{\mathbf{x}} =f(x_1,x_2,...,x_n)\]
.
Then  
\[d \omega^0 =\frac{\partial f}{\partial x_1} dx_1 +...+ \frac{\partial f}{\partial x_n} dx_n \]

and
\[\begin{equation} \begin{aligned} d(d \omega^0) &= (\frac{\partial^2 f}{\partial x_1 \partial x_1} dx_1 +...+ \frac{\partial^2 f}{\partial x_n \partial x_1} dx_n) \wedge dx_1 \\ &+ (\frac{\partial^2 f}{\partial x_1 \partial x_2} dx_1 +...+ \frac{\partial^2 f}{\partial x_n \partial x_2} dx_n) \wedge dx_2 +... \\ &+ (\frac{\partial^2 f}{\partial x_1 \partial x_n} dx_1 +...+ \frac{\partial^2 f}{\partial^2 x_n \partial x_n} dx_n) \wedge dx_n \\ &= \sum^n_{i,j=1} \frac{\partial^2 f}{\partial x_i \partial x_j} dx_i ) \wedge dx_j \\&= \sum^n_{i,j=1, \: i j} \frac{\partial^2 f}{\partial x_i \partial x_j} dx_i \wedge dx_j \\ &= \sum^n_{i,j=1, \: i i} \frac{\partial^2 f}{\partial x_i \partial x_j} dx_j \wedge dx_i \\ &= \sum^n_{i,j=1, \: i