Proving $\sum_{k=a}^{n+a}(-1)^{k-a}*k^{b}*{n \choose k - a}==0\enspace\forall\enspace a,b,n>0\enspace and\enspace b$\sum_{k=a}^{n+a}(-1)^{k-a}*k^{b}*{n \choose k - a}==0\enspace\forall\enspace a,b,n>0\enspace and\enspace b<n$ Is this always true and why?
Full story:
I was trying to prove that if you have a polynomial function of degree n and evaluate the function at n + 2 points with a constant step. i.e: evaluate at X0, X0 + 1 * step, X0 + 2 * step, ...., X0 + (n - 1) * step. And then take the difference of the results and then repeat the same step n + 1 times you will always get a zero. 
Example
 A: Proof by induction over the degree of the function $n$ is probably the easiest way.
Base case when $n=0$, $f(x)=c$ so $f(x)-f(x-1)=c-c=0$ takes only 1 step.
Now assumes the result holds for all $f(x)$ of degree $n$. We look at the $n+1$ degree case.
Let $f(x)=a_{n+1}x^{n+1}+a_nx^n+...+a_1x+a_0$. Then
$$g(x)=f(x)-f(x-1)=a_{n+1}(x^{n+1}-(x-1)^{n+1})+h(x)$$
where $h(x)$ has degree at most $n$.
We use binomial theorem to expand $(x-1)^{n+1}=x^{n+1}-...$ so $x^{n+1}-(x-1)^{n+1}$ has degree $n$.
So $g(x)$ has degree $n$ and will take $n+1$ steps to reach $0$.
Now add the extra step for computing $g(x)$ we take $n+2$ steps for $f(x)$ to reach $0$ and by induction we have proven the result.
A: (Nothing original here)
$\sum_{k=a}^{n+a}(-1)^{k-a}k^{b}{n \choose k - a}
=\sum_{k=0}^{n}(-1)^{k}(k+a)^{b}{n \choose k}
$.
This is the
$n$-th difference of
$f(x)
=(x+a)^b
$,
which is a polynomial
of degree $b$.
Since the first difference
($f(x)-f(x-1)$)
of a polynomial of
degree $b$
is a polynomial of degree
$b-1$,
the $n$-th difference,
up to $n = b$,
is a polynomial of
degree $b-n$.
Therefore
the $b$-th difference
of a polynomial of
degree $b$
is a constant,
so all higher differences
are zero.
