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?
 A: You can continue by changing the order of summation and then splitting the sum into two parts:
\begin{align*}
& \sum\limits_{n = 0}^\infty  {\frac{{( - 1)^{n + 1} \pi ^{2n + 1} }}{{(2n + 1)!}}\left( {\sum\limits_{k = 0}^{2n + 1} {( - 1)^k \binom{2n + 1}{k}(z + 1)^k } } \right)} 
\\ & = \sum\limits_{k = 0}^\infty  {\left( {\sum\limits_{n = \left\lceil {\frac{{k - 1}}{2}} \right\rceil }^\infty  {\frac{{( - 1)^{n + k + 1} \pi ^{2n - k + 1} }}{{(2n - k + 1)!}}} } \right)\frac{{\pi ^k }}{{k!}}(z + 1)^k } 
\\ & = \sum\limits_{k = 0}^\infty  {\left( {\sum\limits_{n = k}^\infty  {\frac{{( - 1)^{n + 2k + 1} \pi ^{2n - 2k + 1} }}{{(2n - 2k + 1)!}}} } \right)\frac{{\pi ^{2k} }}{{(2k)!}}(z + 1)^{2k} } \\ & \quad + \sum\limits_{k = 0}^\infty  {\left( {\sum\limits_{n = k}^\infty  {\frac{{( - 1)^{n + 2k + 2} \pi ^{2n - 2k} }}{{(2n - 2k)!}}} } \right)\frac{{\pi ^{2k + 1} }}{{(2k + 1)!}}(z + 1)^{2k + 1} } 
\\ & = \sum\limits_{k = 0}^\infty  {\left( {( - 1)^{k + 1} \sum\limits_{j = 0}^\infty  {\frac{{( - 1)^j \pi ^{2j + 1} }}{{(2j + 1)!}}} } \right)\frac{{\pi ^{2k} }}{{(2k)!}}(z + 1)^{2k} }  \\ & \quad + \sum\limits_{k = 0}^\infty  {\left( {( - 1)^k \sum\limits_{j = 0}^\infty  {\frac{{( - 1)^j \pi ^{2j} }}{{(2j)!}}} } \right)\frac{{\pi ^{2k + 1} }}{{(2k + 1)!}}(z + 1)^{2k + 1} } 
\\ & = \sum\limits_{k = 0}^\infty  {\left( {( - 1)^{k + 1} \sin \pi } \right)\frac{{\pi ^{2k} }}{{(2k)!}}(z + 1)^{2k} }  + \sum\limits_{k = 0}^\infty  {\left( {( - 1)^k \cos \pi } \right)\frac{{\pi ^{2k + 1} }}{{(2k + 1)!}}(z + 1)^{2k + 1} } \\ & = \sum\limits_{k = 0}^\infty  {( - 1)^{k + 1} \frac{{\pi ^{2k + 1} }}{{(2k + 1)!}}(z + 1)^{2k + 1} } .
\end{align*}
A: 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.
