$(\nabla\cdot\mathbf{n})^{2}+(\mathbf{n}\times\nabla\times\mathbf{n})^{2}+(\mathbf{n}\cdot\nabla\times\mathbf{n})^{2}=|\nabla\bf n|^2$

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I would like to show the following (we are in 2D, we apply Einstein's notation, and the norm is defined by $|n|=n_in_i$)

$$\mathbf{n} = n_x \mathbf{e}_x+n_y\mathbf{e}_y$$

$$(\nabla\cdot\mathbf{n})^{2}+(\mathbf{n}\times\nabla\times\mathbf{n})^{2}+(\mathbf{n}\cdot\nabla\times \mathbf{n})^{2}=|\nabla \mathbf{n}|^2:=\sum_{i=1}^2\sum_{j=1}^2(\partial_i n_j)^2$$

where $|\mathbf{n}|=1$, $\nabla\times\bf n=\partial_x n_y-\partial_y n_x$ and I didn't use Einstein's notation for the last term, in order to emphasise how it is to be intended.

what would be the quickest way possible, using Einstein's index notations or ...?