Prove that $\sum_{X=0}^N u(X) {N \choose X} p^X (1-p)^{N-X}=0 \iff u(X)=0, \space \forall X\in\{ 0,1,...,N \}$ I am trying to prove that $Bin(N,p)$ where $N$ is fixed is a complete distribution.
Thus my goal is to show 
$$E[u(X)]=0 \iff u(X)=0$$ 
While I was attempting to prove this I have noticed that 
$$\sum_{X=0}^N u(X) {N \choose X} p^X (1-p)^{N-X}$$
is a degree-$N$ polynomial congruent to $0$ making all coefficients equal to $0$.
Here, the coefficients ends up being a nice linear combination which I suspect that it is a form of binomial coefficients.
For example, when $N=3$ I get the following
$$\begin{align} \sum_{X=0}^3 u(X) {3 \choose X} p^X (1-p)^{3-X}= \\
& \quad p^3*(u(3)-3u(2)+3u(1)-u(0)) \\ 
&+p^2*3(u(2)-2u(1)+u(0)) \\
&+p*3(u(1)-u(0)) \\
&+1(u(0))\\
\end{align}$$
$$ = p^3\sum_{i=0}^3u(i){3 \choose i}(-1)^i +p^2\sum_{i=0}^2u(i){2 \choose i}(-1)^i +p\sum_{i=0}^1u(i){1 \choose i}(-1)^i + u(0)$$
$$=\sum_{j=0}^3\sum_{i=0}^j u(i){3 \choose j}{j \choose i}(-1)^ip^j$$
The part that I would like have assistance is to show that 
$$\sum_{X=0}^N u(X) {N \choose X} p^X (1-p)^{N-X}=\sum_{j=0}^N\sum_{i=0}^j u(i){N \choose j}{j \choose i}(-1)^ip^j$$
and that the cascades of $u(i)=0$ occurs, i.e., 
 $$u(0)=0 \implies u(1)=0 \implies ... \implies u(N)=0$$
I appreciate your assistance.

This is an answer that summarizes the question and the comments.
The goal is to show that 
$$E[u(X)]=0 \iff u(X)=0$$
and the given equation is equivalent to 
$$\sum_{X=0}^n u(X) {N \choose X} p^X (1-p)^{N-X} = \sum_{j=0}^N \sum_{i=0}^j u(i){N \choose j} {j \choose i}(-1)^ip^j = 0$$
We are assuming that $N$ is fixed and $p \in [0,1]$
The middle column is a Bernstein Polynomial that is a base, thus if it is equal to $0$ then $u(X)=0$.
This shows that $Bin(N,p)$ is a complete distribution.
