Let $f(t) = \sum_{i = 0}^{n} c_i \ t^i$ be a degree $n$ polynomial. In the binomial basis, we can write $f(t) = \sum_{k = 0}^{n} h_k \ \binom{t + n - k}{n}$. Using the calculus of finite differences, I can prove that the coefficient $c_i$ can be computed from $\{ h_0, \dots, h_n \}$ by the equation $c_i = \frac{1}{n!} \sum_{k = 0}^{n} \left( \sum_{\ell = i}^{n} s(n,\ell) \binom{\ell}{i} (n-k)^{\ell - i} \right) h_k, $ where $s(n,\ell)$ is the (signed) Stirling Number of the first kind. I'd like to have the inverse formula. That is, how does one compute $h_k$ as a function of $\{ c_0, \dots, c_n \}$?
Thanks.
Update: Thanks to the suggestions below, I can prove \begin{eqnarray} f(t) = \sum_{i = 0}^{n} \sum_{k = 0}^{i} \sum_{j = k}^{i} c_{i} k! \binom{i}{j} r^{i-j} S(j,k) \binom{t - r}{k} \end{eqnarray} for any positive integer $r$. I'm not sure how to convert this into an expansion in the basis $\left \{ \binom{t + n - k}{k} \right\}$ or $\left\{ \binom{t + n -k}{n} \right\}$, since I believe that $r$ cannot depend on $k$. Any ideas are certainly appreciated!