I am closely following Hamilton's Mathematical Gauge Theory.
Let $V$ be a vector space, $Q$ a bilinear form on $V$, and $CL(V,Q)$ the corresponding Clifford algebra. We can construct the Clifford algebra by taking the quotient of the tensor algebra of $V$, denoted $T(V)$, and the ideal $I(V)$ that is generated by the set $\{v \otimes v + Q(v,v)\cdot 1 : v \in v\}$.
Implicit in the definition of the Clifford map is a map $\gamma: V \rightarrow CL(V,Q)$. We may define the map $\gamma$ as $\gamma = \pi \circ i$ where $i: V \rightarrow T(V)$ is the inclusion map and $\pi: T(V) \rightarrow Cl(V,Q) = T(V)/I(V)$ the quotient map.
We use the notation $\mathbb{C}l(d)$ being the Clifford algebra $(V, Q) = (\mathbb{C}^d, q)$ where $q$ is the standard non-degenerate complex bilinear form.
Furthermore, let $Cl^0(V,Q)$ be the even part of the Clifford algebra, which is defined as $$CL^0(V,Q) = T^0(V)\big/\big(T^0(V) \cap I(Q))\big)$$ where $T^0(V)$ is the subspace of $T(V)$ consisting of tensors with even degree.
Then Hamilton gives the following lemma:
For any $n \geq 1$, $$\mathbb{C}l^0(n) \cong \mathbb{C}l(n-1).$$
In his proof he lets $e_n$ be the $n$-th standard basis vector of $\mathbb{C}^n$ and defines the map $$\delta: \mathbb{C}^{n-1} \rightarrow \mathbb{C}l^0(n)\\ x \mapsto x\cdot e_n$$ and claims that $$\delta(x)^2 = x\cdot e_n \cdot x \cdot e_n = x \cdot x = -q(x,x).$$
I am new to Clifford algebras and I am having trouble doing actual computations with them. How does it follow that $x\cdot e_n \cdot x \cdot e_n = x \cdot x$? Here Hamilton is identifying the products with their images under the map $\gamma$.
Clifford algebras are in general not commuative, so I am guessing that $x \cdot e_n = x$, but I cannot prove this.
After some more thought I was able to figure this out. The key idea is to use the fundamental Clifford relation $x \cdot e_n = -e_n \cdot x - 2q(e_n, x)1$. Since $e_n$ is orthogonal to $x$ (because $x$ is given by a linear combination of $e_1, \ldots, e_{n-1}$) it follows that $q(e_n, x) = 0$. Furthermore, $q(e_n, e_n) = 1$ by definition, so that $$x \cdot e_n \cdot x \cdot e_n = x \cdot e_n \cdot \big(-e_n \cdot x - 2q(e_n, x)\big)\\ = -x \cdot e_n \cdot e_n \cdot x\\ = x \cdot x.$$