Falling and rising factorials are defined as $$ \color{blue}{n^{\underline{k}}}=\,\,\color{blue}{\prod_{j=0}^{k-1}(n-j)} \qquad\qquad \color{blue}{n^{\overline{k}}}=\,\,\color{blue}{\prod_{j=0}^{k-1}(n+j)} $$
Show $$(n+m)^{\underline{k}}= \sum_{v=0}^\infty { k \choose v}\cdot {(m)^{\underline{k-v}}} \cdot (n)^{\underline{v}}$$
Maybe inductive proof? please no combinatoric proof
Note that$$n^{\underline{k}}=\prod_{j=0}^{k-1}(n-j) = k! {n \choose k} $$ and ${k \choose v} =0$ for $v >k$. So we have to prove $$k!{n+m \choose k} = \sum_{v=0}^{k}{k \choose v}(k-v)!{m \choose k-v}v!{n \choose v}$$ $$=\sum_{v=0}^{k} k! {m \choose k-v}{n \choose v} $$ That is, we have to show$$ {n+m \choose k} = \sum_{v=0}^{k} {m \choose k-v}{n \choose v} \tag{1}$$ The LHS of $(1)$ is the coefficient of $x^k$ in $(1+x)^{m+n}$ and the RHS is the coefficient of $x^k$ in $(1+x)^m(1+x)^n$, so they are indeed equal and $(1)$ holds.