I am reading a book named "An Introduction to Clifford Algebras and Spinors" by J. Vaz Jr. and R. da Rocha. In page 78, I met a confusing formula (3.89), written as:
$$\gamma(\mathbf{v})\gamma(\mathbf{u})=\gamma[\gamma(\mathbf{v})(\mathbf{u})]$$
where $\gamma:V\to\mathcal{C}\ell(V,g)$ is a Clifford mapping and $\mathbf{v},\mathbf{u}\in V$. The book said this formula is due to "the associativity of Clifford algebra".
I really don't understand this, could anyone give some hints of why?
Context:
In the preceding section the authors constructed Clifford algebra using creation operators $\mathbf{E}:V\to\mathrm{End}(\bigwedge(V))$ where $\mathbf{E}(\mathbf{v})(A)=\mathbf{v}\wedge A$, and annihilation operators $\mathbf{I}:V^*\to\mathrm{End}(\bigwedge(V))$ where $\mathbf{I}(\alpha)(A)=\alpha\rfloor A$. Using notation $\flat:V\to V^*$ as the index lowering operator, the Clifford mapping is defined as
$$\gamma=\gamma_+=\mathbf{E}+\mathbf{I}\circ\flat$$
After that confusing formula, the authors used above formula to get ($g$ is the associated bilinear quadratic form)
$$\begin{eqnarray} \gamma(\mathbf{v})(\mathbf{u})&=&(\mathbf{E}+\mathbf{I}\circ\flat)(\mathbf{v})(\mathbf{u})\\ &=&\mathbf{v}\wedge\mathbf{u}+\mathbf{v}_\flat\rfloor\mathbf{u}\\ &=&\mathbf{v}\wedge\mathbf{u}+g(\mathbf{v},\mathbf{u}) \end{eqnarray}$$
And, thus
$$\begin{eqnarray} \gamma(\mathbf{v})\gamma(\mathbf{u})&=&\gamma[\gamma(\mathbf{v})(\mathbf{u})]\\ &=&\gamma[\mathbf{v}\wedge\mathbf{u}+g(\mathbf{v},\mathbf{u})]\\ &=&\gamma(\mathbf{v}\wedge\mathbf{u})+g(\mathbf{v},\mathbf{u})\mathbf{1} \end{eqnarray}$$
I asked the original author via email and finally understand the formula. So let me share my understanding below.
Suppose we have an associative operator $\vartriangle$, satisfying $$(\mathbf{v}\vartriangle\mathbf{u})\vartriangle\mathbf{w}=\mathbf{v}\vartriangle\mathbf{u}\vartriangle\mathbf{w}=\mathbf{v}\vartriangle(\mathbf{u}\vartriangle\mathbf{w})$$
So here if $\mathbf{v}\vartriangle\mathbf{u}=\gamma(\mathbf{v})(\mathbf{u})$, then we have $$\begin{eqnarray} \gamma[\gamma(\mathbf{v})(\mathbf{u})](\mathbf{w})&=&(\mathbf{v}\vartriangle\mathbf{u})\vartriangle\mathbf{w}\\ &=&\mathbf{v}\vartriangle(\mathbf{u}\vartriangle\mathbf{w})\\ &=&\gamma(\mathbf{v})(\gamma(\mathbf{u})(\mathbf{w}))\\ &=&\gamma(\mathbf{v})\gamma(\mathbf{u})(\mathbf{w}) \end{eqnarray}$$
Removing $\mathbf{w}$ from both sides, we finally get $$\gamma(\mathbf{v})\gamma(\mathbf{u})=\gamma[\gamma(\mathbf{v})(\mathbf{u})]$$
And the associativity is due to Clifford algebra.