Taylor expansion of $\sin \pi z$ at $z = -1$. Taylor expansion of $\sin \pi z$ at $z = -1$ is
$$\sin\pi z = -\sin(\pi(z+1)) = -\sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}(z+1)^{2n+1}$$ so that
$$\sin\pi z = \sum_{n=0}^\infty \frac{(-1)^{n+1}\pi^{2n+1}}{(2n+1)!}(z+1)^{2n+1}. \tag{$\dagger$}$$
But if I try this
\begin{align}
\sin\pi z  & = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}z^{2n+1} \\& = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}(z+1-1)^{2n+1}  \\ & = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}\left(\sum_{k=0}^{2n+1}\binom{2n+1}{k}(z+1)^k(-1)^{2n-k+1}\right).\tag{$\dagger^*$}
\end{align}
In this case, how can I reduce $(\dagger^*)$ as the above form $(\dagger)$? Just direct calculation?

Start from
$$\sin\pi z = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}z^{2n+1}$$
and
$$-1 = \cos\pi = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n}}{(2n)!}$$
Consider
$$\sin\pi z \cos\pi = \sin(\pi (z+1)) = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}z^{2n+1}\left(\sum_{n=0}^\infty \frac{(-1)^n\pi^{2n}}{(2n)!}\right)$$
and compare the coefficients of $z^k$ to
$$\sum_{n=0}^\infty \frac{(-1)^{n+1}\pi^{2n+1}}{(2n+1)!}(z+1)^{2n+1}$$
They are the same.
