The Chevalley–Eilenberg differential squares to zero

Expand $d(df)$ using [L1], and let $X^{ij}$ denote the ordered list obtained by omitting $x_i,x_j$. For fixed $i<j$, the outer action by $x_i$ followed by the action of $x_j$ has coefficient $(-1{)}^{i+j-1}$, whereas the reverse order has coefficient $(-1{)}^{i+j}$. Their sum is $(-1{)}^{i+j-1}[\rho (x_i),\rho (x_j)]f(X^{ij})$. There is exactly one term in which the outer differential forms $[x_i,x_j]$ and the inner differential lets that new first argument act; its coefficient is $(-1{)}^{i+j}$, so it contributes $(-1{)}^{i+j}\rho ([x_i,x_j])f(X^{ij})$. These three terms cancel by [L2].

It remains to account for action--bracket terms on three distinct indices. Fix $i<j$ and $k\mathrm{\notin }\{i,j\}$. Let $a$ be the number of $i,j$ that are less than $k$, and $b$ the number greater than $k$, so $a+b=2$. Acting first by $x_k$ and then forming $[x_i,x_j]$ has coefficient $(-1{)}^{i+j+k-b}$. Forming $[x_i,x_j]$ first and then letting $x_k$ act has coefficient $(-1{)}^{i+j+k+1-a}$. The exponents differ by $1-a+b=3-2a$, which is odd, while both terms have the same value $\rho (x_k)f([x_i,x_j],X^{ijk})$. Thus they cancel. This covers every mixed term with three distinct original indices.

Consider the terms in which both differentials use their bracket sums. For two disjoint pairs, the two possible orders have the same scalar sign: the numbers of cross-pair index shifts in the two orders add to $4$, so their sign exponents differ by an even integer. Their cochain values are $f([x_p,x_q],[x_i,x_j],X^{ijpq})$ and $f([x_i,x_j],[x_p,x_q],X^{ijpq})$, which cancel because $f$ is alternating. For $i<j<k$, the three terms in which the second bracket uses the bracket created by the first have common coefficient $(-1{)}^{i+j+k-1}$ and bracket sum $[[x_i,x_j],x_k]+[[x_j,x_k],x_i]-[[x_i,x_k],x_j]$. Since $-[[x_i,x_k],x_j]=[[x_k,x_i],x_j]$, this is zero by Jacobi [L3]. The expansion has now been partitioned into action--action plus created-bracket action (step 1.1), mixed action--bracket terms (step 1.2), disjoint double brackets, and nested double brackets; hence $d^{2}f=0$. For $n<0$ or a zero cochain space the assertion is the unique zero composite, and for $n=0$ step 1.1 is exactly the representation identity.