Induction proof of $\sum_{j=0}^n{(-1)^j{n \choose j}\prod_{k=m+1}^{m+n-1}{(j+k)}}=0$ Does anynone have some hints to prove the following equation by induction for all $n\geq 1$ and $m\in\mathbb{Z} $
$$\sum_{j=0}^n{(-1)^j{n \choose j}\prod_{k=m+1}^{m+n-1}{(j+k)}}=0$$
use for induction step:
$$ {n-1 \choose k-1} + {n-1 \choose k} = {n \choose k} $$
 A: Start with 
$$\sum_{j=0}^{n} (-1)^j \binom{n}{j} (1+x)^{j} = (-x)^n $$
By applying binomial theorem to $(1 - (1+x))^n$.
Now multiply both sides with $(1+x)^{m+n-1}$ and differentiate ... times, and set $x$ to ..., and show by induction that ...
A: If you're looking for a proof by induction ... this isn't it. Notice
$$\sum_{j=0}^n (-1)^j{n \choose j} \left[\prod_{k=m+1}^{m+n-1}{(j+k)}\right]x^{j+m}  =\sum_{j=0}^{n}(-1)^j{n\choose j}\frac{d^{n-1}}{dx^{n-1}}\left(x^{j+m+n-1}\right)$$
What can be done from here?
$$ \begin{array}{l} =\frac{d^{n-1}}{dx^{n-1}}\left[x^{m+n-1}\sum_{j=0}^n{n\choose j}(-x)^j\right] \\
 =\frac{d^{n-1}}{dx^{n-1}} \left(x^{m+n-1}(1-x)^n\right) \\ 
 =\sum_{l=0}^{n-1}\left(\frac{d^{n-1-l}}{dx^{n-1-l}}x^{m+n-1}\right)\left(\frac{d^{\,l}}{dx^l}(1-x)^n\right).\end{array}$$ Now, $\displaystyle\frac{d^{\,l}}{dx^l}(1-x)^n =(-1)^l \left[\prod_{r=n-l+1}^n r\right](1-x)^{n-l}$ and $(1-x)^{n-l}=0$ at $x=1$ when $l<n$.
A: We can write the product as a factorial times a binomial and then use Vandermonde's Identity to write that as a linear combination of $\binom{j}{i}$:
$$
\begin{align}
\prod_{k=m+1}^{m+n-1}(j+k)
&=(n-1)!\binom{j+m+n-1}{n-1}\\
&=(n-1)!\sum_{i=0}^{n-1}\binom{j}{i}\binom{m+n-1}{n-i-1}\tag{1}
\end{align}
$$
$(1)$ is a degree $n-1$ polynomial in $j$, and in general, the $n^{\text{th}}$ forward difference of a polynomial of degree less than $n$ is $0$. In particular
$$
\begin{align}
&\sum_{j=0}^n(-1)^j\binom{n}{j}\prod_{k=m+1}^{m+n-1}(j+k)\\
&=(-1)^n\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}\prod_{k=m+1}^{m+n-1}(j+k)\tag{2}\\
&=(-1)^n\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}(n-1)!\sum_{i=0}^{n-1}\binom{j}{i}\binom{m+n-1}{n-i-1}\tag{3}\\
&=(-1)^n(n-1)!\sum_{i=0}^{n-1}\binom{m+n-1}{n-i-1}\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}\binom{j}{i}\tag{4}\\
&=(-1)^n(n-1)!\sum_{i=0}^{n-1}\binom{m+n-1}{n-i-1}\sum_{j=0}^n(-1)^{n-j}\binom{n-i}{n-j}\binom{n}{i}\tag{5}\\
&=(-1)^n(n-1)!\sum_{i=0}^{n-1}\binom{m+n-1}{n-i-1}(1-1)^{n-i}\binom{n}{i}\tag{6}\\[12pt]
&=0\tag{7}
\end{align}
$$
Explanation of Steps:
$(2)$: $(-1)^j=(-1)^n(-1)^{n-j}$
$(3)$: apply $(1)$
$(4)$: rearrange terms
$(5)$: $\displaystyle\binom{n}{j}\binom{j}{i}=\frac{n!}{(n-j)!\,i!\,(j-i)!}=\binom{n}{i}\binom{n-i}{n-j}$
$(6)$: $\displaystyle\sum_{j=0}^n(-1)^{n-j}\binom{n-i}{n-j}=(1-1)^{n-i}$
$(7)$: each term in $(6)$ is $0$ since $i\lt n$
