Why is $\operatorname{Div}\big(\operatorname{Curl} F\big) = 0$? Intuition? Why is $\operatorname{Div}\big(\operatorname{Curl} F\big) = 0$? Is there an intuitive explanation to what this means as well as an algebraic proof?

Also I understand that $\operatorname{Curl} \operatorname{Grad}F =0$, I interpet this as the gradient not having any rotation since its the direction of steepest ascent and if this formed some sort of loop this would not make sense? Is this somewhat correct? Any explanations would be appreciated.
Algebraically, both of these are simply based on the equality of mixed partials:
$$
\begin{align}
\nabla\cdot\nabla\times u
&amp;=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\cdot\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\times\left(u_1,u_2,u_3\right)\\
&amp;=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\cdot\left(\frac{\partial u_3}{\partial y}-\frac{\partial u_2}{\partial z},\frac{\partial u_1}{\partial z}-\frac{\partial u_3}{\partial x},\frac{\partial u_2}{\partial x}-\frac{\partial u_1}{\partial y}\right)\\
&amp;=\color{#0000FF}{\frac{\partial^2u_3}{\partial x\partial y}}-\color{#00A000}{\frac{\partial^2u_2}{\partial x\partial z}}+\color{#C00000}{\frac{\partial^2u_1}{\partial y\partial z}}-\color{#0000FF}{\frac{\partial^2u_3}{\partial y\partial x}}+\color{#00A000}{\frac{\partial^2u_2}{\partial z\partial x}}-\color{#C00000}{\frac{\partial^2u_1}{\partial z\partial y}}\\[9pt]
&amp;=0\tag{1}\\[20pt]
\nabla\times\nabla u
&amp;=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\times\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)u\\
&amp;=\left(\frac{\partial^2}{\partial y\partial z}-\frac{\partial^2}{\partial z\partial y},\frac{\partial^2}{\partial z\partial x}-\frac{\partial^2}{\partial x\partial z},\frac{\partial^2}{\partial x\partial y}-\frac{\partial^2}{\partial y\partial x}\right)u\\[6pt]
&amp;=(0,0,0)\,u\tag{2}\\[20pt]
\end{align}
$$
Intuitively(?), $(1)$ says that a flow induced perpendicular to the rotation of a smooth vector field has no source, and $(2)$ says that the flow induced by the gradient of a smooth scalar field has no rotation.
Here's an informal way of interpreting it. Taking the curl of a function is a cross product operation: curl. Now let's try to take the divergence of this curl, however, recall that cross products produce orthogonal vectors. Since the divergence is a dot product operation divergence, and the dot product of orthogonal vectors is 0, the divergence of a curl is 0.