I am trying to solve Whittaker & Watson's Chap 6, Misc Example 6.13. I understand the formula to be $$ \int_0^\infty \left( \frac {\sin t} t \right)^n \,dt = \frac{\pi}{2^n (n-1)!} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n\choose k} (n-2k)^{n-1}. $$
Through a trick, I brought this to an equivalent calculation that the Euclidean volume of a portion of the $(n-1)$-dimensional cube $$ \{(x_1,\cdots,x_{n-1})\,|\, -1\leq x_i\leq 1, -1\leq x_1 + \cdots + x_{n-1}\leq 1\} $$ is $$ \frac{1}{(n-1)!} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n\choose k} (n-2k)^{n-1}. $$
I can check the volume calculation for $n$ up to 4, and $(n-1)$ up to 3, but I don't know how to proceed in general.
Help with either proof would be appreciated, and it would imply the other one.
HINT:
Using the binomial theorem, write
$$\begin{align} \sin^n(z)&=\left(\frac{e^{iz}-e^{-iz}}{2i}\right)^n\\\\ &=\frac1{(2i)^n}\sum_{k=0}^n \binom{n}{k}(-1)^{k}e^{i(n-2k)z} \end{align}$$
Now, apply the residue theorem. Be careful to close the plane in the upper half plane for $n>2k$ and in the lower half plane for $n<2k$.