(1)

\begin{align*} \sum_{k=n}^{m}C\left(m,k\right)C\left(k,n\right) & =\sum_{k=n}^{m}\frac{m!}{k!\left(m-k\right)!}\frac{k!}{n!\left(k-n\right)!}\\ & =\frac{m!}{n!}\sum_{k=n}^{m}\frac{1}{\left(m-k\right)!\left(k-n\right)!}\\ & =\frac{m!}{n!\left(m-n\right)!}\sum_{k=n}^{m}\frac{\left(m-n\right)!}{\left(m-k\right)!\left(k-n\right)!}\\ & =C\left(m,n\right)\sum_{k=n}^{m}C\left(m-n,k-n\right)\\ & =C\left(m,n\right)\sum_{k=0}^{m-n}C\left(m-n,k\right)\\ & =2^{m-n}C\left(m,n\right) \end{align*}

(2)

\begin{align*} \sum_{k=0}^{\infty}\sum_{j=0}^{k-a}\left(-1\right)^{j}C\left(k,j+a\right)C\left(j+b,c\right)t^{k} & =\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\left(-1\right)^{j}C\left(k,j+a\right)C\left(j+b,c\right)t^{k}\\ & =\sum_{j=0}^{\infty}\sum_{k=j}^{\infty}\left(-1\right)^{j}C\left(k,j+a\right)C\left(j+b,c\right)t^{k}\\ & =\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\left(-1\right)^{j}C\left(k+j,j+a\right)C\left(j+b,c\right)t^{k+j}\\ & =\sum_{j=0}^{\infty}C\left(j+b,c\right)\left(-t\right)^{j}\sum_{k=0}^{\infty}C\left(k+j,j+a\right)t^{k}\\ & =\sum_{j=0}^{\infty}C\left(j+b,c\right)\left(-t\right)^{j}\sum_{k=0}^{\infty}C\left(k+j+a,j+a\right)t^{k+a}\cmt{\because0\leq a}\\ & =\sum_{j=0}^{\infty}C\left(j+b,c\right)\left(-t\right)^{j}t^{a}\left(1-t\right)^{-\left(j+a+1\right)}\\ & =\frac{t^{a}}{\left(1-t\right)^{a+1}}\sum_{j=0}^{\infty}C\left(j+b,c\right)\left(\frac{-t}{1-t}\right)^{j}\\ & =\frac{t^{a}}{\left(1-t\right)^{a+1}}\sum_{j=0}^{\infty}C\left(j+c,c\right)\left(\frac{-t}{1-t}\right)^{j}\cmt{\because b\leq c}\\ & =\frac{t^{a}}{\left(1-t\right)^{a+1}}\left(\frac{-t}{1-t}\right)^{c-b}\left(1-t\right)^{c+1}\\ & =\left(-1\right)^{c-b}t^{a-b+c}\left(1-t\right)^{b-a}\\ & =\left(-1\right)^{c-b}t^{a-b+c}\sum_{k=0}^{\infty}C\left(b-a,k\right)\left(-t\right)^{k}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k+c-b}C\left(b-a,k\right)t^{k+a-b+c}\\ & =\sum_{k=a-b+c}^{\infty}\left(-1\right)^{k-a}C\left(b-a,k-a+b-c\right)t^{k}\\ & =\sum_{k=a-b+c}^{\infty}\left(-1\right)^{k-a}C\left(b-a,c-k\right)t^{k} \end{align*} となるので係数を比べると、
\[ \sum_{j=0}^{k-a}\left(-1\right)^{j}C\left(k,j+a\right)C\left(j+b,c\right)=\begin{cases} \left(-1\right)^{k-a}C\left(b-a,c-k\right) & a-b+c\leq k\\ 0 & k<a-b+c \end{cases} \] となる。
例えば\(a=1,b=c=k=0\)や\(a=2,b=c=k=1\)のとき上の式は成り立たない。
従って与式は成り立つ。