(1)

フルヴィッツ・ゼータ関数のハンケル経路積分表示より、
\begin{align*} \zeta\left(-n,\alpha\right) & =-\frac{\Gamma\left(1+n\right)}{2\pi i}\int_{C}\frac{\left(-z\right)^{-\left(n+1\right)}e^{-\alpha z}}{1-e^{-z}}dz\\ & =\frac{\Gamma\left(1+n\right)}{2\pi i}\int_{C}\left(-z\right)^{-\left(n+2\right)}\frac{z}{e^{z}-1}e^{\left(1-\alpha\right)z}dz\\ & =\frac{\Gamma\left(1+n\right)}{2\pi i}\int_{C}\left(-z\right)^{-\left(n+2\right)}\left(\sum_{k=0}^{\infty}\frac{B_{k}}{k!}z^{k}\right)\left(\sum_{j=0}^{\infty}\frac{\left(1-\alpha\right)^{j}}{j!}z^{j}\right)dz\\ & =\frac{\Gamma\left(1+n\right)}{2\pi i}\int_{C}\left(-z\right)^{-\left(n+2\right)}\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\frac{B_{k}\left(1-\alpha\right)^{j}}{k!j!}z^{k+j}dz\\ & =\frac{\Gamma\left(1+n\right)}{2\pi i}\int_{C}\left(-z\right)^{-\left(n+2\right)}\sum_{k=0}^{\infty}\sum_{j=0}^{k}\frac{B_{k-j}\left(1-\alpha\right)^{j}}{\left(k-j\right)!j!}z^{k}dz\\ & =\left(-1\right)^{n}\frac{\Gamma\left(1+n\right)}{2\pi i}\sum_{k=0}^{\infty}\sum_{j=0}^{k}\int_{C}\frac{B_{k-j}\left(1-\alpha\right)^{j}}{\left(k-j\right)!j!}z^{-n+k-2}dz\\ & =\left(-1\right)^{n}\frac{\Gamma\left(1+n\right)}{2\pi i}\sum_{k=0}^{\infty}\sum_{j=0}^{k}2\pi i\delta_{k,n+1}\frac{B_{k-j}\left(1-\alpha\right)^{j}}{\left(k-j\right)!j!}\\ & =\left(-1\right)^{n}\frac{\left(1+n\right)!}{n+1}\sum_{j=0}^{n+1}\frac{B_{n+1-j}\left(1-\alpha\right)^{j}}{\left(n+1-j\right)!j!}\\ & =\left(-1\right)^{n}\frac{1}{n+1}\sum_{j=0}^{n+1}C\left(n+1,j\right)B_{n+1-j}\left(1-\alpha\right)^{j}\\ & =\left(-1\right)^{n}\frac{1}{n+1}B_{n+1}\left(1-\alpha\right)\\ & =-\frac{1}{n+1}B_{n+1}\left(\alpha\right) \end{align*}

(2)

\begin{align*} \zeta\left(-n\right) & =\zeta\left(-n,1\right)\\ & =-\frac{1}{n+1}B_{n+1}\left(1\right) \end{align*}