= Solution
Use the complex <inner product> convention linear in its first argument. Give the <exterior algebra> its induced <inner product>, with an <orthonormal> wedge basis. Exterior multiplication is \b[$e(v)\eta=v\wedge\eta$]. Its adjoint is contraction:
$$
e(b)^*(v_1\wedge\cdots\wedge v_r)
=\sum_{j=1}^r(-1)^{j-1}(v_j,b)\,v_1\wedge\cdots\widehat v_j\cdots\wedge v_r.
$$
Applying this formula to $a\wedge\eta$ separates its first term from the remaining terms, giving $e(b)^*e(a)\eta=(a,b)\eta-e(a)e(b)^*\eta$. Hence the <canonical anticommutation relations> are
$$
\boxed{e(a)e(b)^*+e(b)^*e(a)=(a,b)I.}
$$
For irreducibility, choose an <orthonormal basis> $u_1,\ldots,u_d$. The occupancy projections are $N_j=e(u_j)e(u_j)^*$, and $P_0=\prod_j(I-N_j)$ projects onto the vacuum $1\in\Lambda^0V$. For an increasing index set $J$, put $E_J=e(u_{j_1})\cdots e(u_{j_r})$. Then $E_IP_0E_J^*$ maps the wedge basis <vector> $u_J$ to $u_I$ and kills every other wedge basis <vector>. Thus these products are all <matrix units> and the generated star-algebra is \b[$\operatorname{End}(\Lambda V)$]. Any common <invariant subspace> for the <creation and annihilation operators> is therefore either zero or all of $W$.
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