Setup
I've been reading/looking into variations on the $si(n) := \int_0^{\infty}\textrm{sinc}^n(x)$ for some power $n$ an integer, or the more general $si(a,b):= \int_0^{\infty}\frac{\sin^a(x)}{x^b}$ (a formula for $si(a,b)$ is provided on this page of Wolfram Encyclopedia with a reference to an article that is lost to history, but see [1] of [2] for a proof of this formula).
Anyway, solutions to $si(n)$ have been asked a number of times (e.g., [3], [4], [5], and others I'm sure), but I was interested in the oft-pointed to paper Some Remarkable Properties of Sinc and Related Integrals by Borwein and Borwein, which uses Fourier transforms to derive the solution.
My Question: Combinatorial Identity in Borwein and Borwein
In the Borwein paper, they define the following in Section 3, Remarks 1c (I've simplified for my relevant question): $$\gamma = (\gamma_1,\cdots,\gamma_n)\in\{-1,1\}^n,\quad \epsilon_{\gamma} = \prod_{i=1}^n\gamma_i,\quad b_{\gamma} = 1+\sum_{i=1}^n\gamma_i$$ They claim the following identity:
\begin{equation} \sum_{\gamma\in\{-1,1\}^{n-1}, b_{\gamma}<0}\epsilon_{\gamma}b_{\gamma}^{n-1} = \sum_{1\leq r\leq \frac{n}{2}} (-1)^{r+1}\binom{n-1}{r-1}(n-2r)^{n-1}\tag{1} \end{equation}
Equation (1) helps define the solution $si(n)$.
How to prove (1) (see edit)
Here's my short attempt so far: Let's call $c_{\gamma}:=b_{\gamma}-1 = \sum_{i=1}^{n-1}\gamma_i<-1$ This only occurs when the number of $+1's$ in $\gamma$, called $s$ satisfies $0\leq s< \frac{n-1}{2}$. My combinatorics is weak, but pretty sure the number of $\gamma$ such that $c_{\gamma}=-k$ is something like $\binom{n-1}{\frac{n-1-k}{2}}$. We can expand $b_{\gamma}^{n-1}$ using the binomial theorem: $$ b_{\gamma}^{n-1} = \left(1+c_{\gamma}\right)^{n-1} = \sum_{j=0}^{n-1}\binom{n-1}{j}c_{\gamma}^{n-1-j}\tag{2} $$
I assume that I can just plug (2) back into (1) and do some simple algebra (while counting the number of ways for $b_{\gamma}$ or $c_{\gamma}$ to equal some value) but it appears not to be working, but I'll work more on it and edit if I make progress.
Edit:
So now I think I understand how to show this and I may answer my own question later when I have a chance to sit down and work this out. Counting the number of $\pm 1's$ is more complicated than this needs to be and is probably not going to help, instead reindex by the values of $b_{\gamma}$ which are in $\{2-n,4-n,\cdots,-2\textrm{ or } -1\}$. So we divide into cases when $b_{\gamma} = -(n-2r)$ for $1\leq r\leq \lfloor \frac{n}{2}\rfloor$. Then (1) becomes: $$ \sum_{\gamma\in\{-1,1\}^{n-1}, b_{\gamma}<0}\epsilon_{\gamma}b_{\gamma}^{n-1} = \sum_{1\leq r\leq \lfloor \frac{n}{2}\rfloor}(-1)^{n-1}\epsilon_{\gamma}C_{\gamma}(n-2r)^{n-1}\tag{3} $$
For $C_{\gamma}$ the number of binary vectors $\gamma$ such that $b_{\gamma} = -(n-2r)$. This now seems pretty simple: prove that $C_{\gamma} = \binom{n-1}{r-1}$, which is probably pretty easy, and $(-1)^{n-1}\epsilon_{\gamma} = (-1)^{r+1}$, which is also probably easy since it is equivalent to $\epsilon_{\gamma} = (-1)^{r+1 - (n -1)} = (-1)^{r-n}$, so there will be some manipulation here.
