An alternating sum I ran into an alternating sum in my research and would like to know if the following identity is true:
$$
\sum_{i = 0}^{\left\lfloor \left(n + 1\right)/2\right\rfloor} \frac{\left(n + 1 - 2i\right)^{n + 1}}{2^{n}\left(n + 1\right)!}\binom{n + 1}{i}\left(-1\right)^{i} = 1\quad
\forall\ \mbox{positive integers}\ n\geq 3
$$
Any help would be appreciated!.
Edit.
We might try to use an Iverson bracket $[[2q\le n]]$ in attempting to
evaluate
$$S_n = \sum_{q=0}^{\lfloor n/2\rfloor}
(n-2q)^n {n\choose q} (-1)^q.$$
We obtain
$$[v^n] \frac{1}{1-v}
\sum_{q\ge 0} v^{2q} (n-2q)^n {n\choose q} (-1)^q$$
Using a coefficient extractor,
$$n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{n+1}} \frac{1}{1-v}
\sum_{q\ge 0} v^{2q} \exp(-2qz) {n\choose q} (-1)^q
\\ = n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{n+1}} \frac{1}{1-v} (1-v^2\exp(-2z))^n.$$
Now residues sum to zero and the residue at one yields
$$- n! [z^n] \exp(nz) (1-\exp(-2z))^n.$$
We have that since $(1-\exp(-2z))^n = (2z-2z^2\pm\cdots)^n = 2^n z^n +\cdots$ this evaluates to $-2^n n!.$ We find for the residue at infinity
$$- n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^2}
v^{n+1} \frac{1}{1-1/v} (1-\exp(-2z)/v^2)^n
\\ = n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^n} \frac{1}{1-v} (v^2-\exp(-2z))^n
\\ = n! [z^n] \exp(nz) \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^n} \frac{1}{1-v}
\sum_{q=0}^n {n\choose q} (-1)^{n-q} \exp(-2(n-q)z) v^{2q}
\\ = \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^n} \frac{1}{1-v}
\sum_{q=0}^n {n\choose q} (-1)^{n-q} (2q-n)^n v^{2q}
\\ = \sum_{q=0}^n {n\choose q} (-1)^q (n-2q)^n [[2q\le n-1]].$$
Now when $n$ is odd this gives the upper limit $\lfloor n/2\rfloor$ and
when $n$ is even $\lfloor n/2\rfloor -1$ however in the latter case we
may raise to $\lfloor n/2\rfloor$ because the added term is zero in the
sum per $(n-2q)^n = 0$. We have obtained
$$\sum_{q=0}^{\lfloor n/2\rfloor} {n\choose q} (-1)^q (n-2q)^n
= S_n.$$
Collecting everything we have shown that $S_n - 2^n n! + S_n = 0$ or
$S_n = 2^{n-1} n!.$ The question now becomes, is there a simpler proof?
 A: We seek to show that
$$\sum_{q=0}^{\lfloor n/2\rfloor}
(n-2q)^n {n\choose q} (-1)^q
= 2^{n-1} n!.$$
Observe that
$$(n-2q)^n {n\choose q} (-1)^q
= \frac{1}{2}
\left[ (n-2q)^n {n\choose q} (-1)^q
+ (2q-n)^n {n\choose n-q} (-1)^{n-q} \right].$$
Hence we get for our sum
$$\frac{1}{2}
\sum_{q=0}^{\lfloor n/2\rfloor}
(n-2q)^n {n\choose q} (-1)^q
+ \frac{1}{2}
\sum_{q=0}^{\lfloor n/2\rfloor}
(2q-n)^n {n\choose n-q} (-1)^{n-q}
\\ = \frac{1}{2}
\sum_{q=0}^{\lfloor n/2\rfloor}
(n-2q)^n {n\choose q} (-1)^q
+ \frac{1}{2}
\sum_{q=n-\lfloor n/2\rfloor}^n
(n-2q)^n {n\choose q} (-1)^q
\\ = \frac{1}{2}
\sum_{q=0}^n
(n-2q)^n {n\choose q} (-1)^q.$$
Introducing a coefficient extractor,
$$\frac{1}{2} \sum_{q=0}^n {n\choose q} (-1)^q
n! [z^n] \exp((n-2q) z)
\\ = \frac{1}{2} n! [z^n] \exp(nz)
\sum_{q=0}^n {n\choose q} (-1)^q
\exp((-2q) z)
\\ = \frac{1}{2} n! [z^n] \exp(nz)
(1-\exp(-2z))^n.$$
Note however that $(1-\exp(-2z))^n = (2z-2z^2\pm\cdots)^n$ so the only
contribution to the coefficient extractor $[z^n]$ is from the first
term of the series so that $[z^n] \exp(nz) (2z-2z^2\pm\cdots)^n = 2^n$ and we
finally have
$$2^{n-1} n!$$
as claimed.
