(1)

ベルヌーイ関数母関数より、
\begin{align*} \sum_{k=0}^{\infty}B_{k}\left(-x\right)\frac{t^{k}}{k!} & =\frac{te^{-xt}}{e^{t}-1}\\ & =\frac{te^{x\left(-t\right)}e^{-t}}{1-e^{-t}}\\ & =\frac{\left(-t\right)e^{x\left(-t\right)}}{e^{-t}-1}e^{-t}\\ & =\sum_{k=0}^{\infty}B_{k}\left(x\right)\frac{\left(-t\right)^{k}}{k!}e^{-t}\\ & =\sum_{k=0}^{\infty}B_{k}\left(x\right)\frac{\left(-t\right)^{k}}{k!}\sum_{j=0}^{\infty}\frac{\left(-t\right)^{j}}{j!}\\ & =\sum_{k=0}^{\infty}\sum_{j=0}^{k}B_{k-j}\left(x\right)\frac{\left(-t\right)^{k}}{\left(k-j\right)!j!}\\ & =\sum_{k=0}^{\infty}\frac{t^{k}}{k!}\left(-1\right)^{k}\sum_{j=0}^{k}B_{k-j}\left(x\right)C\left(k,j\right) \end{align*} となり、両辺の係数を比べて、
\begin{align*} \left(-1\right)^{k}B_{k}\left(-x\right) & =\sum_{j=0}^{k}C\left(k,j\right)B_{k-j}\left(x\right)\\ & =\sum_{j=0}^{k}C\left(k,k-j\right)B_{j}\left(x\right)\\ & =\sum_{j=0}^{k}C\left(k,j\right)B_{j}\left(x\right) \end{align*} となる。
従って与式は成り立つ。

(2)

(1)より、
\begin{align*} \sum_{k=0}^{n}C\left(n+1,k\right)B_{k}\left(x\right) & =\sum_{k=0}^{n+1}C\left(n+1,k\right)B_{k}\left(x\right)-C\left(n+1,n+1\right)B_{n+1}\left(x\right)\\ & =\left(-1\right)^{n+1}B_{n+1}\left(-x\right)-B_{n+1}\left(x\right)\\ & =B_{n+1}\left(1+x\right)-B_{n+1}\left(x\right)\\ & =\left(n+1\right)x^{n} \end{align*}

(2)-2

\begin{align*} \sum_{k=0}^{n}C\left(n+1,k\right)B_{k}\left(x\right) & =\sum_{k=0}^{n}C\left(n+1,k\right)\sum_{j=0}^{k}C\left(k,j\right)B_{j}x^{k-j}\\ & =\sum_{j=0}^{n}\sum_{k=j}^{n}C\left(n+1,k\right)C\left(k,j\right)B_{j}x^{k-j}\\ & =\sum_{j=0}^{n}B_{j}\sum_{k=j}^{n}\frac{\left(n+1\right)!}{k!\left(n+1-k\right)!}\cdot\frac{k!}{j!\left(k-j\right)!}x^{k-j}\\ & =\sum_{j=0}^{n}B_{j}\frac{\left(n+1\right)!}{j!\left(n-j+1\right)!}\sum_{k=j}^{n}\frac{\left(n-j+1\right)!}{\left(n+1-k\right)!\left(k-j\right)!}x^{k-j}\\ & =\sum_{j=0}^{n}B_{j}C\left(n+1,j\right)\sum_{k=j}^{n}C\left(n-j+1,k-j\right)x^{k-j}\\ & =\sum_{j=0}^{n}B_{j}C\left(n+1,j\right)\sum_{k=0}^{n-j}C\left(n-j+1,k\right)x^{k}\\ & =\sum_{j=0}^{n}B_{j}C\left(n+1,j\right)\left\{ \sum_{k=0}^{n-j+1}C\left(n-j+1,k\right)x^{k}-C\left(n-j+1,n-j+1\right)x^{n-j+1}\right\} \\ & =\sum_{j=0}^{n}B_{j}C\left(n+1,j\right)\left\{ \left(1+x\right)^{n+1-j}-x^{n+1-j}\right\} \\ & =\sum_{j=0}^{n+1}B_{j}C\left(n+1,j\right)\left\{ \left(1+x\right)^{n+1-j}-x^{n+1-j}\right\} -B_{n+1}\left\{ \left(1+x\right)^{0}-x^{0}\right\} \\ & =\sum_{j=0}^{n+1}B_{j}C\left(n+1,j\right)\left\{ \left(1+x\right)^{n+1-j}-x^{n+1-j}\right\} \\ & =B_{n+1}\left(1+x\right)-B_{n+1}\left(x\right)\\ & =\left(n+1\right)x^{n} \end{align*}

(3)

\begin{align*} \sum_{k=0}^{n-1}C\left(n,k\right)B_{k}\left(x\right) & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}\left(x\right)-C\left(n,n\right)B_{n}\left(x\right)\\ & =\left(-1\right)^{n}B_{n}\left(-x\right)-B_{n}\left(x\right)\\ & =B_{n}\left(1+x\right)-B_{n}\left(x\right)\\ & =nx^{n-1} \end{align*}

(4)

\begin{align*} B_{n}\left(x+y\right) & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}\left(x+y\right)^{n-k}\\ & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}\sum_{j=0}^{n-k}C\left(n-k,j\right)x^{j}y^{n-k-j}\\ & =\sum_{k=0}^{n}C\left(n,k\right)B_{k}\sum_{j=k}^{n}C\left(n-k,j-k\right)x^{j-k}y^{n-j}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{j}C\left(n,k\right)B_{k}C\left(n-k,j-k\right)x^{j-k}y^{n-j}\\ & =\sum_{j=0}^{n}\sum_{k=0}^{j}C\left(j,k\right)C\left(n,j\right)B_{k}x^{j-k}y^{n-j}\\ & =\sum_{j=0}^{n}C\left(n,j\right)y^{n-j}\sum_{k=0}^{j}C\left(j,k\right)B_{k}x^{j-k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)y^{n-j}B_{j}\left(x\right)\\ & =\sum_{j=0}^{n}C\left(n,j\right)B_{j}\left(x\right)y^{n-j} \end{align*}

(5)

略