(1)

カタラン数の母関数より、
\begin{align*} \sum_{k=0}^{\infty}C_{k}x^{k} & =\frac{1-\sqrt{1-4x}}{2x}\\ & =\frac{1}{2x}-\frac{1}{2x}\sum_{k=0}^{\infty}C\left(\frac{1}{2},k\right)\left(-4x\right)^{k}\\ & =\frac{1}{2x}-\frac{1}{2x}\sum_{k=0}^{\infty}\frac{P\left(\frac{1}{2},k\right)}{k!}\left(-4x\right)^{k}\\ & =\frac{1}{2x}-\frac{1}{2x}\left\{ 1+\sum_{k=1}^{\infty}\frac{P\left(\frac{1}{2},k\right)}{k!}\left(-4x\right)^{k}\right\} \\ & =-\frac{1}{2x}\sum_{k=1}^{\infty}\frac{P\left(\frac{1}{2},k\right)}{k!}\left(-4x\right)^{k}\\ & =-\frac{1}{2x}\sum_{k=1}^{\infty}\frac{\frac{1}{2}}{\frac{1}{2}-k}\cdot\frac{P\left(-\frac{1}{2},k\right)}{k!}\left(-4x\right)^{k}\\ & =\frac{1}{2x}\sum_{k=1}^{\infty}\frac{P\left(-\frac{1}{2},k\right)}{\left(2k-1\right)k!}\left(-4x\right)^{k}\\ & =\frac{1}{2x}\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}\left(2k-1\right)!}{\left(2k-1\right)k!2^{2k-1}\left(k-1\right)!}\left(-4x\right)^{k}\cmt{P\left(-\frac{1}{2},n\right)=\frac{\left(-1\right)^{n}\left(2n-1\right)!}{2^{2n-1}\left(n-1\right)!}}\\ & =\frac{1}{2x}\sum_{k=1}^{\infty}\frac{2\left(2k-2\right)!}{k!\left(k-1\right)!}x^{k}\\ & =\sum_{k=1}^{\infty}\frac{\left(2k-2\right)!}{k!\left(k-1\right)!}x^{k-1}\\ & =\sum_{k=0}^{\infty}\frac{\left(2k\right)!}{\left(k+1\right)!k!}x^{k}\\ & =\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{k+1}x^{k} \end{align*} となる。
これより、係数を比較して
\[ C_{k}=\frac{1}{k+1}C\left(2k,k\right) \] となるので与式は成り立つ。

(2)

(1)より、
\begin{align*} C_{n} & =\frac{1}{n+1}C\left(2n,n\right)\\ & =\left(1-\frac{n}{n+1}\right)C\left(2n,n\right)\\ & =C\left(2n,n\right)-\frac{n}{n+1}C\left(2n,n\right)\\ & =C\left(2n,n\right)-C\left(2n,n-1\right) \end{align*} となるので与式は成り立つ。