How to prove a formula involving polynomial sequences and their recursive representation If we have a sequence defined by the polynomial $a_n=\displaystyle \sum_{k=0}^{m}c_kn^k$, then how can we prove that $a_n=\displaystyle \sum_{k=1}^{m+1}\binom{m+1}{k} (-1)^{k-1}a_{n-k}$?
*Edited to fix the typo, and also simplified

The LHS is $$a_n = \sum_{k=0}^m c_k  n^k$$ and the RHS is (typo in the
leading term corrected)
$$(-1)^{m} a_{n-m-1}
+\sum_{p=1}^m {m+1\choose p} (-1)^{p-1} a_{n-p}.$$
Following @MartyCohen we merge these two to get
$$\sum_{p=1}^{m+1} {m+1\choose p} (-1)^{p-1} a_{n-p}.$$
This is
$$\sum_{p=1}^{m+1} {m+1\choose p} (-1)^{p-1} 
\sum_{k=0}^m c_k (n-p)^k
\\ = \sum_{k=0}^m c_k
\sum_{p=1}^{m+1} {m+1\choose p} (-1)^{p-1} (n-p)^k.$$
Working with the inner sum we introduce
$$(n-p)^k =
\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}} \exp((n-p)z) \; dz.$$
This yields for the double sum
$$\sum_{k=0}^m c_k
\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}}
\sum_{p=1}^{m+1} {m+1\choose p} (-1)^{p-1} \exp((n-p)z) \; dz
\\ = -\sum_{k=0}^m c_k
\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}} \exp(nz)
\sum_{p=1}^{m+1} {m+1\choose p} (-1)^{p} \exp(-pz) \; dz.$$
The inner sum is
$$(1-\exp(-z))^{m+1} - 1.$$
We thus require
$$k! [z^k] \exp(nz) (1-\exp(-z))^{m+1} - k! [z^k] \exp(nz).$$
There   are   two  pieces   here.    Since  $1-\exp(-z)   =   z   -
\frac{1}{2}z^2+\cdots$ the exponentiated  component of the first piece
starts at $z^{m+1}.$ But $k\le m$ so we have a contribution of zero.
The second piece yields $$-n^k.$$
The result then becomes
$$\sum_{k=0}^m c_k n^k = a_n$$
which is the claim.
