Let $\mathfrak{g}$ be a Lie algebra and let $x_1, \dotsc, x_n$ be elements of $\mathfrak{g}$. Is there an easy way to see that $$ \sum_{1 \leq i < j \leq n} (-1)^{i+j} [x_i, x_j] \wedge x_1 \wedge \dotsb \wedge \widehat{x_i} \wedge \dotsb \wedge \widehat{x_j} \wedge \dotsb \wedge x_n = 0 $$ whenever $x_k = x_l$ for some $k \neq l$?
So far I’ve only managed to prove this identity by induction over $n$, case distinction and some calculations. But this is neither nice nor convincing.
This identity is used in the constructions of Lie algebra homology to ensure that the differential $\bigwedge^n \mathfrak{g} \otimes M \to \bigwedge^{n-1} \mathfrak{g} \otimes M$ is well-defined.
The map $\mathfrak{g}^n\to\bigwedge^{2}\mathfrak{g}\otimes\bigwedge^{n-2}\mathfrak{g}$ $$\phi:(x_1,\ldots,x_n)\mapsto\sum_{1 \leq i < j \leq n} (-1)^{i+j} (x_i\wedge x_j)\otimes( x_1 \wedge \cdots \wedge \widehat{x_i} \wedge \cdots \wedge \widehat{x_j} \wedge \cdots \wedge x_n)$$ is alternating and multilinear (why?). As the Lie bracket is alternating, it's in effect a map $\bigwedge^2\mathfrak{g}\to\mathfrak{g}$. Composing $\phi$ with this gives map $\mathfrak{g}^n\to\mathfrak{g}\otimes\bigwedge^{n-2}\mathfrak{g}$ and then composing with exterior product gives a map $\mathfrak{g}^n\to\bigwedge^{n-1}\mathfrak{g}$. All these maps are multilinear and alternating on $\mathfrak{g}^n$. Your assertion is that the final map is alternating.
So it boils down to showing that $\phi$ is alternating. One can prove this by the same sort of hacking that you are trying to avoid. But there is a more conceptual way: $\phi$ corresponds to a linear map $\Phi:\bigwedge^n\mathfrak{g}\to\bigwedge^{2}\mathfrak{g}\otimes\bigwedge^{n-2}\mathfrak{g}$ and this is part of a comultiplication map $\Delta:\bigwedge\mathfrak{g} \to \bigwedge\mathfrak{g} \, {\hat{\otimes}} \, \bigwedge\mathfrak{g}$ on the exterior algebra $\bigwedge\mathfrak{g}$. Here $\hat\otimes$ denotes that the multiplication on the tensor product is graded skew-commutative: $(\alpha\otimes\beta)(\gamma\otimes\delta)=(-1)^{rs}(\alpha\gamma\otimes\beta\delta)$ where $\beta$ and $\gamma$ gave degrees $r$ and $s$ respectively.
The construction of $\Delta$ comes from the universal property of the exterior power. We define $\Delta(x)=x\otimes 1+1\otimes x$ for $x\in\mathfrak{g}$ and by skew-commutativity, $\Delta(x)^2=0$. The comultiplication decomposes via the grading as $\Delta_{r,s}:\bigwedge^{r+s}\mathfrak{g}\to\bigwedge^2\mathfrak{g}\otimes\bigwedge^s\mathfrak{g}$, and $\Phi$ is $-\Delta_{2,n-2}$.