Evaluate $$\dfrac{\binom{n}{0}}{m!}+n\cdot\dfrac{\binom{n}{1}}{(m+1)!}+n(n-1)\cdot\dfrac{\binom{n}{2}}{(m+2)!}+\ldots (n \, \, \text{terms})$$
I tried to integrate $x^{m!-1}(1+x)^n$ from $0$ to $1$. But I won't get product of $n$'s in numerator. Any hint?
Consider the following definition, the rising factorial of $x$ is defined as $x^{\overline{n}}=x(x+1)\cdots (x+n-1),$ and the falling factorial is defined as $x^{\underline{n}}=x(x-1)\cdots (x-n+1).$ Your expression can be put in those terms as $$\sum _{k=0}^n\binom{n}{k}\frac{n^{\underline{k}}}{(m+k)!}=\color{red}{\frac{(m+n)!}{(m+n)!}}\sum _{k=0}^n\binom{n}{k}\frac{n^{\underline{k}}}{(m+k)!}=\frac{1}{(m+n)!}\sum _{k=0}^n\binom{n}{k}{n^{\underline{k}}\cdot (m+n)^{\underline{n-k}}}$$ There is a binomial theorem of the falling factorial, which is: $$(a+b)^{\underline{n}}=\sum _{k=0}^n\binom{n}{k}a^{\underline{k}}b^{\underline{n-k}},$$ so your expression ends up being $$\frac{(m+2n)^{\underline{n}}}{(m+n)!}=\frac{n!}{(m+n)!}\binom{m+2n}{n}$$