Let $V$ be a vector space with some extra structure (maybe to be general, an object of an abelian tensor category?), where we can form tensor products and exterior powers $\Lambda^i V$ and symmetric powers $\text{Sym}^i V$. I believe the following always holds for $n > 0$:
$$ \sum\_{i=0}^n (-1)^i\Lambda^i V \otimes \text{Sym}^{n-i}V = 0$$
where I take the sum as a "virtual" object; another way to say this is just that
$$ \sum\_{i=0}^{[n/2]} \Lambda^{2i} V \otimes \text{Sym}^{n-2i}V = \sum\_{i=0}^{[(n-1)/2]} \Lambda^{2i+1} V \otimes \text{Sym}^{n-2i-1}V.$$


For example, if $V$ is a representation of a finite group, you can use character theory to reduce to the analogous identity for the elementary symmetric and complete homogeneous polynomials (see [Wikipedia](https://en.wikipedia.org/wiki/Complete_homogeneous_symmetric_polynomial#Relation_with_the_elementary_symmetric_polynomials)) applied to the eigenvalues of a group element. But it'd be nice to have a higher-level proof that applied more generally.



Alternating sum of symmetric and exterior powers vanishes
If $V$ is a vector space, then you always have a Koszul complex


$$\cdots \to \bigwedge^i V \otimes Sym(V) \to \bigwedge^{i-1} V \otimes Sym(V) \to \cdots \to V \otimes Sym(V) \to Sym(V)$$


which has no homology except at the end where it's a copy of the ground field in degree $0$ (this complex is naturally graded where $$\bigwedge^i V \otimes Sym^k V$$ has degree $i+k$).


The identity you mention is the Euler characteristic of the degree $n$ piece being $0$. The Koszul complex is equivariant for the natural action of $GL(V)$. If $V$ happens to have more structure, great: it will likely be compatible, for example a finite group acting on $V$ gives a subgroup of $GL(V)$.


