To show that an algebra constructed as a quotient of the tensor algebra of a vector space is nonzero, one of the main ways to go is to construct representations. We can do this for the Clifford algebra as follows.
Let $V$ be a vector space over a field $k$ and $(,):V\times V \to k$ a symmetric bilinear form on $V$. The Clifford algebra (for this form) is given by
$$Cl(V)= T(V)/\langle v \otimes v - (v,v)\rangle.$$
We will construct a representation of the Clifford algebra on the exterior algebra $\bigwedge (V)$.
For $v \in V$, define two $k$-endomorphisms of $\bigwedge(V)$ by
$$ l_v(x) = v \wedge x$$
and
$$ \delta_v(x) = \sum_{j=1}^k (-1)^{k-1}(v,x_j) -1)^{j-1}(v,x_j) x_1 \wedge \dots \wedge \widehat{x_j} \wedge \dots \wedge x_k$$
if $x = x_1 \wedge \dots \wedge x_k$.
Then check that $l_v^2 = \delta_v^2 = 0$, and moreover that $l_v \delta_v + \delta_v l_v = (v,v) \cdot \mathrm{id}$.
Extend the map linear $v \mapsto l_v + \delta_v$ to an algebra homomorphism from the tensor algebra $T(V)$ to $\mathrm{End}_k(\bigwedge(V))$. By the previous remark, this descends to a map, let's call it $\phi$, from the Clifford algebra to $\mathrm{End}_k(\bigwedge(V))$.
In particular, $\phi(v)1 = v$, so $V$ injects into the Clifford algebra.
Edit: I believe also that the map $$ x \mapsto \phi(x)1$$ gives a linear isomorphism of the Clifford algebra with the exterior algebra.
A great reference for this stuff is Chevalley's monograph, The Algebraic Theory of Clifford Algebras and Spinors.
