combinatorics - combinatorial proof for binomial identity

Consider the following binomial identity:
$$\sum_{k=0}^n(-1)^k\binom{n}{k}g(k)=0$$
for every polynomial $g(k)$ with degree less than $n$.

My Proof
Every polynomial $g(k)$ of degree $t$ can be represented in the following form
$$g(k) = \sum_{l=0}^tc_l(k)_l,$$
where
$$(k)_l=k(k-1)\ldots(k-l+1),$$
ans $c_l$ are some coefficients.

For every $l $$\begin{multline} \sum_{k=0}^n(-1)^k\binom{n}{k}(k)_l= \sum_{k=l}^n(-1)^k\binom{n}{k}(k)_l= \sum_{k=l}^n(-1)^k\frac{n!k!}{k!(n-k)!(k-l)!}= \sum_{k=l}^n(-1)^k\frac{n!k!}{k!(n-k)!(k-l)!}= \frac{n!}{(n-l)!}\sum_{k=l}^n(-1)^k\frac{(n-l)!}{(n-k)!(k-l)!}=\\ \frac{n!}{(n-l)!}\sum_{k=l}^n(-1)^k\binom{n-l}{n-k}=0, \end{multline}$$
therefore,
$$\sum_{k=0}^n(-1)^k\binom{n}{k}g(k)=0.$$

Question
Do you know some other proofs of this identity? I'm most interested in combinatorial proof.

Answer

Let $n$ a positive integer and $E$ the real vector space of polynomials with degree $

Consider the linear map $\varphi:E\to E,P(X)\mapsto P(X)-P(X+1)$.

For every nonzero $P(X)\in E$, we observe that $\deg(\varphi(P(X))<\deg(P(X))$. This leads to $\varphi^n=0$ (of course $\varphi^k$ means $\varphi\circ\cdots\circ\varphi$ with $k$ times $\varphi$).

Now consider the map $\psi:E\to E,P(X)\mapsto P(X+1)$, so that $\varphi=id_E-\psi$.

Since $\psi$ and $id_E$ commute, we can apply Newton's binomial formula and get :

$$\sum_{k=0}^n{n\choose k}(-1)^k\psi^k=\left(id_E-\psi\right)^n=\varphi^n=0$$

So, for any $P(X)\in E$ :

$$\sum_{k=0}^n{n\choose k}(-1)^kP(X+k)=0$$ Evaluating in $0$, we finally get :

$$\sum_{k=0}^n{n\choose k}(-1)^kP(k)=0$$