(0)

\begin{align*} \frac{e^{D}-1}{D}B_{n}\left(x\right) & =\sum_{k=1}^{\infty}\frac{D^{k-1}}{k!}B_{n}\left(x\right)\\ & =\sum_{k=1}^{\infty}\frac{P\left(n,k-1\right)}{k!}B_{n-\left(k-1\right)}\left(x\right)\\ & =\frac{1}{n+1}\sum_{k=1}^{\infty}\frac{P\left(n+1,k\right)}{k!}B_{n+1-k}\left(x\right)\\ & =\frac{1}{n+1}\sum_{k=1}^{n+1}C\left(n+1,k\right)B_{n+1-k}\left(x\right)\\ & =\frac{1}{n+1}\sum_{k=0}^{n}C\left(n+1,k\right)B_{k}\left(x\right)\\ & =\frac{1}{n+1}\left(n+1\right)x^{n}\\ & =x^{n} \end{align*} これより、両辺に\(\frac{D}{e^{D}-1}\)を掛けると、
\begin{align*} B_{n}\left(x\right) & =\frac{D}{e^{D}-1}x^{n} \end{align*} となり、与式は成り立つ。

(0)-2

\begin{align*} \frac{e^{D}-1}{D}B_{n}\left(x\right) & =\sum_{k=1}^{\infty}\frac{D^{k-1}}{k!}B_{n}\left(x\right)\\ & =\sum_{k=1}^{\infty}\frac{D^{k-1}}{k!}\sum_{j=0}^{n}C\left(n,j\right)B_{j}x^{n-j}\\ & =\sum_{k=1}^{\infty}\frac{1}{k!}\sum_{j=0}^{n}C\left(n,j\right)B_{j}P\left(n-j,k-1\right)x^{n-j-\left(k-1\right)}\\ & =\sum_{j=0}^{n}C\left(n,j\right)B_{j}x^{n-j}\sum_{k=0}^{\infty}\frac{P\left(n-j,k\right)}{\left(k+1\right)!}\left(\frac{1}{x}\right)^{k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)B_{j}x^{n-j}\frac{1}{n-j+1}\sum_{k=0}^{\infty}\frac{P\left(n-j+1,k+1\right)}{\left(k+1\right)!}\left(\frac{1}{x}\right)^{k}\\ & =\sum_{j=0}^{n}C\left(n,j\right)B_{j}x^{n-j+1}\frac{1}{n-j+1}\sum_{k=0}^{\infty}C\left(n-j+1,k+1\right)\left(\frac{1}{x}\right)^{k+1}\\ & =\sum_{j=0}^{n}C\left(n,j\right)B_{j}x^{n-j+1}\frac{1}{n-j+1}\left\{ \sum_{k=0}^{\infty}C\left(n-j+1,k\right)\left(\frac{1}{x}\right)^{k}-C\left(n-j+1,0\right)\left(\frac{1}{x}\right)^{0}\right\} \\ & =\sum_{j=0}^{n}C\left(n,j\right)B_{j}x^{n-j+1}\frac{1}{n-j+1}\left\{ \left(1+\frac{1}{x}\right)^{n-j+1}-1\right\} \\ & =\frac{1}{n+1}\sum_{j=0}^{n}C\left(n+1,j\right)B_{j}x^{n-j+1}\left\{ \left(\frac{1+x}{x}\right)^{n-j+1}-1\right\} \\ & =\frac{1}{n+1}\sum_{j=0}^{n}C\left(n+1,j\right)B_{j}\left\{ \left(1+x\right)^{n-j+1}-x^{n-j+1}\right\} \\ & =\frac{1}{n+1}\left\{ \sum_{j=0}^{n+1}C\left(n+1,j\right)B_{j}\left\{ \left(1+x\right)^{n+1-j}-x^{n+1-j}\right\} -C\left(n+1,n+1\right)B_{n+1}\left\{ \left(1+x\right)^{0}-x^{0}\right\} \right\} \\ & =\frac{1}{n+1}\sum_{j=0}^{n+1}C\left(n+1,j\right)B_{j}\left\{ \left(1+x\right)^{n+1-j}-x^{n+1-j}\right\} \\ & =\frac{1}{n+1}\left(B_{n+1}\left(1+x\right)-B_{n+1}\left(x\right)\right)\\ & =\frac{1}{n+1}\left(n+1\right)x^{n}\\ & =x^{n} \end{align*} これより、両辺に\(\frac{D}{e^{D}-1}\)を掛けると、
\begin{align*} B_{n}\left(x\right) & =\frac{D}{e^{D}-1}x^{n} \end{align*} となり、与式は成り立つ。