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 $\gamma = 1$.