Prove that $\nabla\times(\nabla\phi)=\mathbf{0}$, with $\phi$ a scalar field.
I tried to write out all the terms, what gave me the following (where $i$,$j$,$k$ is the standard basis for $\mathbb{R}^3$): \begin{align*} \nabla\times(\nabla\phi)=&\;\nabla\times\left(\frac{\partial \phi}{\partial x}\textbf{i}+\frac{\partial \phi}{\partial y}\textbf{j}+\frac{\partial \phi}{\partial z}\textbf{k}\right)\\ =&\left(\frac{\partial }{\partial x}\textbf{i}+\frac{\partial }{\partial y}\textbf{j}+\frac{\partial }{\partial z}\textbf{k}\right)\left(\frac{\partial \phi}{\partial x}\textbf{i}+\frac{\partial \phi}{\partial y}\textbf{j}+\frac{\partial \phi}{\partial z}\textbf{k}\right)\\ =&\;\frac{\partial^2 \phi}{\partial x^2}\textbf{i}+\frac{\partial^2 \phi}{\partial y^2}\textbf{j}+\frac{\partial^2 \phi}{\partial z^2}\textbf{k}\\ =&\;\nabla^2\phi \end{align*}
But right now, im stuck. I've gained the Laplacian operator for phi, but why is this zero. Have i made a mistake? Can someone please help me?
Using Laplace's rule for determinants of order $3$ or Sarrus' rule, we obtain:
\begin{align} \nabla\times(\nabla\phi)=& \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \dfrac{\partial \phi}{\partial x}& \dfrac{\partial \phi}{\partial y} & \dfrac{\partial \phi}{\partial z} \end{vmatrix}=\\ \left(\dfrac{\partial^2 \phi}{\partial y\partial z}-\dfrac{\partial^2 \phi}{\partial y\partial z}\right)\mathbf{i}-\left(\dfrac{\partial^2 \phi}{\partial x\partial z}-\dfrac{\partial^2 \phi}{\partial x\partial z}\right)\mathbf{j}+\left(\dfrac{\partial^2 \phi}{\partial x\partial y}-\dfrac{\partial^2 \phi}{\partial x\partial y}\right)\mathbf{k}=\mathbf{0} \end{align}
Remember that I have used the Schwarz's theorem and inside the round brackets we have null quantities.