For every Lie algebra $\mathfrak{g}$ we can consider the Chevalley-Eilenberg complex given by the exterior powers $\bigwedge^n \mathfrak{g}$ together with the differential $d_{\mathrm{CE}} \colon \bigwedge^n \mathfrak{g} \to \bigwedge^{n-1} \mathfrak{g}$ that is given by $$ d_{\mathrm{CE}}(x_1 \wedge \dotsb \wedge x_n) = \sum_{1 \leq i < j \leq n} (-1)^{i+j-1} [x_i, x_j] \wedge x_1 \wedge \dotsb \wedge \widehat{x_i} \wedge \dotsb \wedge \widehat{x_j} \wedge \dotsb \wedge x_n \,. $$ We may regard the Lie algebra $\mathfrak{g}$ as a graded vector space concentrated in degree $1$. Then the tensor algebra $\operatorname{T}(\mathfrak{g})$ becomes a graded Hopf algebra (such that $\mathfrak{g}$ consists of primitive elements) and its quotient $\bigwedge \mathfrak{g}$ inherits the structure of a graded Hopf algebra. If I understand the nlab correctly then the Chevalley-Eilenberg differential $d_{\mathrm{CE}}$ is the unique extension of the Lie bracket $[-,-] \colon \bigwedge^2 \mathfrak{g} \to \mathfrak{g}$ to a graded coderivation for the underlying graded coalgebra structure of $\bigwedge \mathfrak{g}$.
How can we show that $d_{\mathrm{CE}}$ is a graded coderivation of $\bigwedge \mathfrak{g}$, and that is it the unique one that extends the Lie bracket?
I have put my (so far unsuccessful) attempt below.
Coderivation
The differential $d_{\mathrm{CE}}$ is of degree $-1$ so we need to show that \begin{equation} \Delta( d_{\mathrm{CE}}(c) ) = \sum_{(c)} d_{\mathrm{CE}}( c_{(1)} ) \otimes c_{(2)} + (-1)^{|c_{(1)}|} c_{(1)} \otimes d_{\mathrm{CE}}(c_{(2)}) \tag{1} \end{equation} for all $c = x_1 \wedge \dotsb \wedge x_n$ with $x_i \in \mathfrak{g}$ (i.e. that $\Delta$ is a homomorphism of chain complexes with respect to $d_{\mathrm{CE}}$). I started explicitely calculating the right hand side: By writing $x_1 \dotsm x_n$ instead of $x_1 \wedge \dotsb \wedge x_n$ we have \begin{align*} {}& \Delta(x_1 \dotsm x_n) \\ ={}& (x_1 \otimes 1 + 1 \otimes x_1) \dotsm (x_n \otimes 1 + 1 \otimes x_n) \\ ={}& \sum_{k=0}^n \; \sum_{\sigma \in S(k,n-k)} (-1)^{n_k(\sigma)} x_{\sigma(1)} \dotsm x_{\sigma(k)} \otimes x_{\sigma(k+1)} \dotsm x_{\sigma(n)} \end{align*} where $S(k,n-k) \subseteq S_n$ denotes the set of $k$-$(n-k)$-shuffles and $$ n_k(\sigma) = \# \{ (i,j) \mid 1 \leq i \leq k, \, k+1 \leq j \leq n, \, \sigma(i) > \sigma(j) \} \,. $$ For the right hand side of $(1)$ we hence get \begin{align*} {}& \sum_{k=0}^n \; \sum_{\sigma \in S(k,n-k)} (-1)^{n_k(\sigma)} \biggl[ d_{\mathrm{CE}}(x_{\sigma(1)} \dotsm x_{\sigma(k)}) \otimes x_{\sigma(k+1)} \dotsm x_{\sigma(n)} \\ {}& \phantom{ \sum_{k=0}^n \; \sum_{\sigma \in S(k,n-k)} (-1)^{n_k(\sigma)} \biggl[ } + (-1)^k x_{\sigma(1)} \dotsm x_{\sigma(k)} \otimes d_{\mathrm{CE}}(x_{\sigma(k+1)} \dotsm x_{\sigma(n)}) \biggr] \\ ={}& \sum_{k=0}^n \; \sum_{\sigma \in S(k,n-k)} (-1)^{n_k(\sigma)} \\ {}& \biggl[ \biggl( \sum_{1 \leq i < j \leq k} (-1)^{i+j-1} [x_{\sigma(i)}, x_{\sigma(j)}] x_{\sigma(1)} \dotsb \widehat{x_{\sigma(i)}} \dotsb \widehat{x_{\sigma(j)}} \dotsb x_{\sigma(k)} \biggr) \otimes x_{\sigma(k+1)} \dotsm x_{\sigma(n)} \\ {}& + (-1)^k x_{\sigma(1)} \dotsm x_{\sigma(k)} \\ {}& \otimes \biggl( \sum_{1 \leq i < j \leq n-k} (-1)^{i+j-1} [x_{\sigma(k+i)}, x_{\sigma(k+j)}] x_{\sigma(k+1)} \dotsb \widehat{x_{\sigma(k+i)}} \dotsb \widehat{x_{\sigma(k+j)}} \dotsb x_{\sigma(n)} \biggr) \biggr] \,. \end{align*} For the left side of $(1)$ one similarly find that \begin{align*} {}& \Delta( d( x_1 \dotsm x_n) ) \\ ={}& \sum_{1 \leq i < j \leq n} (-1)^{i+j-1} \Delta( [x_i, x_j] x_1 \dotsm \widehat{x_i} \dotsm \widehat{x_j} \dotsm x_n ) \\ ={}& \sum_{1 \leq i < j \leq n} (-1)^{i+j-1} ([x_i, x_j] \otimes 1 + 1 \otimes [x_i, x_j]) (x_1 \otimes 1 + 1 \otimes x_1) \\ {}& \phantom{ \sum_{1 \leq i < j \leq n} (-1)^{i+j-1} } \dotsm (x_i \otimes 1 + 1 \otimes x_i)^{\wedge} \dotsm (x_j \otimes 1 + 1 \otimes x_j)^{\wedge} \dotsm (x_n \otimes 1 + 1 \otimes x_n) \end{align*} I guess that it could make sense that after multiplying out these terms and rearranging the resulting summands both sides of (1) give the same result But I don’t really see this (and even less so for the signs).
Uniqueness
I think I have checked that if $C$ is any graded coalgebra and $d_1, d_2 \colon C \to C$ are graded derivations of the same degree $k$ that their equalizer $$ E := \{ x \in C \mid d_1(x) = d_2(x) \} $$ is then a graded subcoalgebra of $C$. We would like that any coderivation of degree $-1$ of $\bigwedge \mathfrak{g}$ is uniquely determined by the restriction $\bigwedge^2 \mathfrak{g} \to \mathfrak{g}$, so I suspect that $\bigwedge \mathfrak{g}$ is generated as a (graded) coalgebra by its homogeneous parts $\bigwedge^2 \mathfrak{g}$. But I don’t know how to show this, or if this is correct.
I guess that for finite dimensional $\mathfrak{g}$ one could also dualize the problem and consider instead the cochain complex $\bigwedge \mathfrak{g}^*$ together with the (co)differential $d_{\mathrm{CE}}^*$ induced by $d_{\mathrm{CE}}$. Then $\bigwedge \mathfrak{g}^*$ should be a graded algebra with derivation $d_{\mathrm{CE}}^*$, so $d_{\mathrm{CE}}^*$ should be uniquely determined by the restriction $\mathfrak{g}^* \to \bigwedge^2 \mathfrak{g}^*$ because $\bigwedge \mathfrak{g}^*$ is generated as an algebra by $\mathfrak{g}$. But I’m not sure if this actually works.