Descents and excedances are equidistributed on $S_n$

Define $\tau :=c\circ \sigma \circ c$. Then, for every $i\in n$, $\tau (i)<i$ if and only if $c(\sigma (c(i)))<i$, which is equivalent to $\sigma (c(i))>c(i)$. Thus $i$ is an anti-excedance of $\tau$ exactly when $c(i)$ is an excedance of $\sigma$, so $\mathrm{\mid }\mathrm{AExc}(\tau )\mathrm{\mid }=\mathrm{exc}(\sigma )$.

The map $\sigma \mapsto \tau =c\circ \sigma \circ c$ is a bijection of $S_n$, because conjugation by a permutation has inverse itself. Also $\mathcal{F}$ is a bijection: from a one-line word, insert a left parenthesis before each left-to-right maximum and a closing parenthesis just before the next such maximum, or at the end, to recover the standard cycle form.

By [L1], $\mathrm{des}(\mathcal{F}(\tau ))=\mathrm{\mid }\mathrm{AExc}(\tau )\mathrm{\mid }$, which equals $\mathrm{exc}(\sigma )$ by step 1.1. Since step 1.2 makes $\sigma \mapsto \mathcal{F}(c\sigma c)$ a bijection of $S_n$, the statistics $\mathrm{des}$ and $\mathrm{exc}$ are equidistributed.