TITLE: Do you know an elegant proof for this expression for a Schur function?
QUESTION [14 upvotes]: I know that the identity
$$
s_\mu = \sum_{\mu-\lambda \text{ is a horizontal strip}} \;\sum_{\alpha\vdash|\lambda|} \frac{\chi^\lambda_\alpha}{z_\alpha} \prod_i(p_i-1)^{a_i}
$$
holds.
Here $\alpha=1^{a_1}2^{a_2}\dotsb$ in exponential notation; in other words, $a_i$ is the number of times that $i$ occurs in $\alpha$.
I know this because I can show that it is essentially equivalent to the formula I.5 of Garsia and Goupil EJC 2009.
Note that the equality of top-degree terms is the well-known expansion of a Schur function in terms of power sum symmetric functions, so it suffices to show that all lower degree terms vanish.
I feel that his simple-looking formula should have a direct and elegant proof. Is there one?

REPLY [6 votes]: Here is another argument. Let $\chi^{\mu/k}$ denote the skew character
of the symmetric group $\mathfrak{S}_n$ corresponding to the skew
shape $\mu/k$. Then by Pieri's rule,
$$ \sum_{\mu-\lambda\ \mathrm{is\ a\ horizontal\ strip}}
      \sum_{\alpha\vdash|\lambda|}\frac{\chi^\lambda_\alpha}{z_\alpha}
      \prod_i p_i^{a_i}
 = \sum_k \sum_{\alpha\vdash n-k}
      \frac{\chi^{\mu/k}_\alpha}{z_\alpha}p_\alpha $$
$$ \qquad = \sum_k s_{\mu/k}. $$
Let $s_k^\perp$ denote the linear operator taking $s_\lambda$ to
$s_{\lambda/k}$. Let $\psi$ denote the linear operator taking $f$ to
$\sum_k s_k^\perp f$. Let $\theta$ denote the
algebra automorphism taking $p_i$ to $p_i-1$.
I claim that $\psi$ and $\theta$ are inverses (so in fact $\psi$ is an
algebra automorphism). By linearity it suffices to show that
$$ \psi p_\lambda = \prod_i (p_{\lambda_i}+1).\ \ \ (*) $$
Now it is not hard to see that $s_k^\perp p_\lambda$ is equal to $\sum
p_\nu$, where $\nu$ is obtained from $\lambda$ by removing a set of
parts (regarding equal parts as distinguishable) summing to $k$. From
this (*) is immediate.
We therefore have
$$ \left.\sum_k s_{\mu/k}\right|_{p_i\to p_i-1} = \theta\psi s_\mu =
    s_\mu, $$
as desired.