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.
 A: 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.
A: What you can do is differentiate with respect to $p$.
If you assume that:
$$\forall p \in [0,1]  \:  \sum_{X=0}^N u(X) \binom N X p^X (1-p)^{N-X} $$
then you certainly have for $p=0$ that $u(0)=0$. So your first term disappears.
When you differentiate once, only the term $X=1$ where you differentiate $p^1$ once remains (you don't need to differentiate the $(1-p)^{N-1}$ because it's multiplied by $p^1$ which will give $0$ when you take $p=0$). So $u(1)=0$ and the second term disappears, and so on and so forth.
What you need to prove to make it work is that:
$$\frac{d^X}{dp^X}\left[p^X(1-p)^{N-X}\right]|_{p=0}=X!$$
and for all $Y>X$:
$$\frac{d^X}{dp^X}\left[p^Y(1-p)^{N-Y}\right]|_{p=0}=0$$
You can use Leibniz formula for a clean proof.
A: Here's an elementary proof.
$\begin{array}\\
s(n, p)
&=\sum_{k=0}^n u(k) {n \choose k} p^k (1-p)^{n-k}\\
&=\sum_{k=0}^n u(n-k) {n \choose n-k} p^{n-k} (1-p)^{k}\\
&=\sum_{k=0}^n \sum_{j=0}^{k}u(n-k) {n \choose k} p^{n-j} \binom{k}{j}(-1)^{k-j}\\
&=\sum_{j=0}^n \sum_{k=j}^{n}u(n-k) {n \choose k} p^{n-j} \binom{k}{j}(-1)^{k-j}\\
&=\sum_{j=0}^n p^j\sum_{k=n-j}^{n}u(n-k) {n \choose k}  \binom{k}{n-j}(-1)^{k-n+j}\\
&=\sum_{j=0}^n p^j(-1)^{n-j}\sum_{k=n-j}^{n}u(n-k) \dfrac{n!k!}{k!(n-k)!(n-j)!(k-n+j)!}(-1)^{k}\\
&=\sum_{j=0}^n p^j(-1)^{j}\dfrac{n!}{(n-j)!}\sum_{k=n-j}^{n}u(n-k)(-1)^{n-k} \dfrac{1}{(n-k)!(k-n+j)!}\\
&=\sum_{j=0}^n p^j(-1)^{j}\dfrac{n!}{(n-j)!}\sum_{k=0}^{j}u(k)(-1)^{k} \dfrac{1}{(k)!(j-k)!}\\
&=\sum_{j=0}^n p^j(-1)^{j}\dfrac{n!}{j!(n-j)!}\sum_{k=0}^{j}u(k)(-1)^{k} \dfrac{j!}{(k)!(j-k)!}\\
&=\sum_{j=0}^n p^j(-1)^{j}\binom{n}{j}\sum_{k=0}^{j}u(k)(-1)^{k}\binom{j}{k}\\
\end{array}
$
If $s(n, p) = 0$
for all $p$,
then,
since it is a polynomial in $p$,
all its coefficients 
must be zero.
Therefore
$\sum_{k=0}^{j}u(k)(-1)^{k}\binom{j}{k} = 0
$
for all $j$.
Starting with $j=0$,
this shows that
$u(k) = 0$
for all $k$.
