## Background

I came up with this while trying to find a sort of high-level exposition of the exterior algebra of a vector space. Let $V$ be a vector space of dimension $n$ over $\mathbb{C}$, and let $k \in \mathbb{N}$. One picture of $\Lambda^k(V)$, the $k^{th}$ exterior power of $V$, is as the space of totally antisymmetric tensors in $V^{\otimes k}$.

This can be constructed as follows. Let $$ \rho : S_k \to \mathrm{End}(V^{\otimes k})$$ be the representation given by $$ \rho_\pi (v_1 \otimes \dots \otimes v_k) = v_{\pi(1)} \otimes \dots \otimes v_{\pi(k)},$$ and then let $\sigma$ be the alternating form of this representation, i.e. $\sigma_\pi = sgn(\pi) \rho_\pi$. The total antisymmetrizer is the map $$A_k = \frac{1}{k!} \sum_{\pi \in S_k} \sigma_\pi.$$ This is the projection onto the space of totally antisymmetric tensors, and so we can calculate the dimension of $\Lambda^k(V)$ simply by taking the trace of the map $A_k$. It turns out that $$\mathrm{tr}(\rho_\pi) = n^{cyc(\pi)},$$ where by $cyc(\pi)$ I mean the number of cycles in the factorization of $\pi$ into disjoint cycles (including cycles of length 1). This can be shown as follows. Take a basis $\{ e_1, \dots, e_n \}$ of $V$ and then form the basis for $V^{\otimes k}$ consisting of all vectors $ e_{i_1} \otimes \dots \otimes e_{i_k}$ such that $1 \le i_1, \dots, i_k \le n $. Then $$ \rho_\pi(e_{i_1} \otimes \dots \otimes e_{i_k}) = e_{i_{\pi(1)}} \otimes \dots \otimes e_{i_{\pi(k)}},$$ so this basis vector contributes 1 to the trace of $\rho_\pi$ if and only if $i_j = i_{\pi(j)}$ for all $1 \le j \le k$, i.e. if and only if all labels are constant over cycles of $\pi$. Since there are $n$ choices for each label, this gives $$\mathrm{tr}(\rho_\pi) = n^{cyc (\pi)},$$ and thus $$ \mathrm{tr}(A_k) = \frac{1}{k!} \sum_{\pi \in S_k} sgn(\pi) n^{cyc (\pi)}.$$

Question: does anybody know a simple combinatorial proof that $$ \frac{1}{k!} \sum_{\pi \in S_k} sgn(\pi) n^{cyc (\pi)} = \binom{n}{k},$$ where (in case you didn't read the long-winded background that I wrote), $cyc(\pi)$ is the number of cycles in the disjoint cycle factorization of $\pi$.