A question about a polynomial 
Suppose that $p$ is a real polynomial of degree $n$. Prove that for $|x|<1$,
  $$\sum\limits_{m=0}^\infty{p(m)x^m}=h((1-x)^{-1})$$
  for some  real polynomial $h$ of degree $n+1$ without the constant term.

I have started trying by taking a specific expression of the polynomial $p$, but I have completely lost. I understand that this result generalizes the following identity
$$(1-x)^{-1}=\sum\limits_{m=0}^\infty{x^m}$$ for $|x|<1$ but couldn't anymore rather thinking about it. Help is highly appreciated.

Consider the polynomials 
$$P_0:=1,P_1:=X, P_2:=X(X-1),\dots,P_l:=X(X-1)\dots (X-l+1).$$
Then we may express $p$ as $\sum_{j=0}^nc_jP_j$ for some coefficients $c_j$. Since for $|x|\lt 1$, we have 
$$\sum_{m=0}^{+\infty}P_j(m)x^m=\frac{d^j}{dx^j}\left(\frac 1{1-x} \right)  $$
and the later can be expressed as a constant times $(1-x)^{-j-1}$, we get the wanted polynomial $h$ by summing over $j$.
