Let $\mathbb{S}_m$ the symmetric group on $m$ letters. Let $v\in\mathbb{S}_m$, and consider paths in the Bruhat order like this: $1\lessdot v_1\lessdot\cdots\lessdot v$, where $\lessdot$ means the covering relation in the (strong) Bruhat order. Let $N_v$ be the number of such paths.

It is intuitively clear that $N_v\leq\ell(v)!$ (for a proof, I found just now the reference), and further that the difference $\ell(v)!-N_v$ is even. Can you prove the latter fact?

**Remark.** What I said should be true for every finite Coxeter group but I am mostly interested in the symmetric group for now.