Proof of $BAC-CAB$ identity missing step I'm stuck on one step of the proof for the identity: $$ \vec{A}\times(\vec{B}\times\vec{C}) = \vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B})$$
So far, the proof follows as:
We know that $\vec{B}\times\vec{C}$ gives a vector perpendicular to both $\vec{B}$ & $\vec{C}$, and that $ \vec{A}\times(\vec{B}\times\vec{C})$ gives a vector perpendicular to both $\vec{A}$ & $(\vec{B}\times\vec{C})$. Therefore, the vector $\vec{A}\times(\vec{B}\times\vec{C})$ must lie in the plane containing both $\vec{B}$ & $\vec{C}$.
Provided $\vec{B}$ & $\vec{C}$ are not parallel (if they were, $\vec{A}\times(\vec{B}\times\vec{C}) = 0$ regardless), vectors $\vec{B}$ & $\vec{C}$ span the 2D plane containing them both.
Therefore, we can  express any vector in the plane as a linear combination of both $\vec{B}$ and $\vec{C}$ and so we can write: $$\vec{A}\times(\vec{B}\times\vec{C}) = \alpha\vec{B} + \beta\vec{C} \tag{1}$$
Taking the scalar product of both sides with $\vec{A}$: $$\vec{A} \cdot (\vec{A}\times(\vec{B}\times\vec{C})) = \vec{A} \cdot (\alpha\vec{B} + \beta\vec{C}) = 0$$ So, $$\alpha(\vec{A} \cdot \vec{B}) + \beta(\vec{A} \cdot\vec{C}) = 0$$
Now writing, $$\lambda = \frac{\alpha}{\vec{A} \cdot\vec{C}} = -\frac{\beta}{\vec{A} \cdot \vec{B}}$$
and substituting $\alpha$ and $\beta$ back into (1) we get:
$$ \vec{A}\times(\vec{B}\times\vec{C}) = \lambda(\vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B}))$$
I am able to show $\lambda = 1$ with particular choices of unit vectors for $\vec{A}, \vec{B}, \vec{C}$ but I am unable to prove that $\lambda$ is independent of the magnitude of vectors (i.e. $\lambda = 1$ for all choices of $\vec{A}, \vec{B}, \vec{C}$). This is the step that I am struggling with. Any suggestions?

(This is a long comment.) This is probably not the simplest way to derive the identity, but the identity follows quite quickly from a special case of Cauchy-Binet formula:
\begin{aligned}
\left(A\times(B\times C)\right)\cdot x
&=\det(A,B\times C,x)\quad\text{(defintion of cross product)}\\
&=\det(x,A,B\times C)\\
&=(x\times A)\cdot(B\times C)\quad\text{(defintion of cross product)}\\
&=(x\cdot B)(A\cdot C)-(x\cdot C)(A\cdot B)\quad\text{(Cauchy-Binet formula)}\\
&=\left(B(A\cdot C)-C(A\cdot B)\right)\cdot x.\\
\end{aligned}
Since $x$ is arbitrary, the result follows.
