Here's another approach. $\def\VA{{\bf A}} \def\VB{{\bf B}} \def\VC{{\bf C}} \def\n{{\bf n}} \def\a{\alpha} \def\b{\beta} \def\g{\gamma}$
We assume $\VA \ne {\bf 0}$ and that $\VA$ and $\n$ are not parallel (or antiparallel) so the vectors $\{\n, \n\times\VA, (\n\times\VA)\times \n\}$ form an orthogonal basis. Then $\begin{align*} \VA &= \a \n + \b \n\times\VA + \g(\n\times\VA)\times\n. \tag{1} \end{align*}$ Using the triple product identity $\VA\times(\VB\times\VC) = \VB(\VA\cdot\VC) - \VC(\VA\cdot\VB)$ we find $\VA = \a \n + \b \n\times\VA + \g[\VA - (\n\cdot\VA)\n].$ We must have $\gamma = 1$, $\alpha = (\n\cdot\VA)$, and $\beta = 0$ so $\VA = (\VA\cdot\n) \n + (\n\times\VA)\times\n$ as claimed.
We can also find the coefficients of (1) in the usual way,
$\begin{align*} \a &= \n\cdot\VA \\ \b &= \frac{(\n\times\VA)\cdot \VA}{(\n\times\VA)^2} \\ \g &= \frac{[(\n\times\VA)\times\n]\cdot \VA}{[(\n\times\VA)\times\n]^2}. \end{align*}$ We immediately have $\beta = 0$. Using the triple product identity we find $\begin{align*} [(\n\times\VA)\times\n]\cdot \VA &= -[\n(\n\cdot\VA)-\VA(\n\cdot\n)]\cdot \VA \\ &= \VA^2 - (\n\cdot\VA)^2 \\ [(\n\times\VA)\times\n]^2 &= [\n(\n\cdot\VA)-\VA(\n\cdot\n)]\cdot[\n(\n\cdot\VA)-\VA(\n\cdot\n)] \\ &= (\n\cdot\VA)^2 - 2(\n\cdot\VA)^2 +\VA^2 \\ &= \VA^2 - (\n\cdot\VA)^2, \end{align*}$ so $\gamma = 1$.