Here's the algebraic proof: not sure if your question is looking for it.
It follows immediately from the definition of curl and div.
In cartesian coordinates for example the curl operator acting on a scalar function is
$curl F = (\partial /\partial y - \partial /\partial z), (\partial /\partial z - \partial /\partial x), (\partial /\partial x - \partial /\partial y) F $
The div operator acting on a vector function G is
$ div G = (\partial /\partial x , \partial /\partial y, \partial / \partial z).G$
So that div.curl acting on a scalar function H is
$div.curl H = (\partial^2 /\partial x.\partial y - \partial^2 / \partial x. \partial z + \partial^2 /\partial y.\partial z - \partial^2 /\partial y.\partial x + \partial ^2 /\partial z . \partial x - \partial^2 /\partial z.\partial y) H $
All you need to know then if that the partial derivative $\partial^2 /\partial x.\partial y $ is equal to $\partial^2 /\partial y.\partial x $ and you can then see that all the terms cancel out leaving zero.