Show that $\nabla m_t$ is a polynomial o degree $p-1$ 
If $m_t=\sum_{k=0}^p c_kt^k$ $t=0,\pm 1,\pm 2,\dots$ show that $\nabla
 m_t$ is a polynomial o degree $p-1$ int $t$ and hence that
  $\nabla^{p+1} m_t=0$

EDIT: I made a mistake $\nabla$ is not a gradient operator, it is a difference operator, such that $\nabla x_t=x_t-x_{
t-1}$
$$\nabla m_t=\sum_{k=0}^p c_kt^k-\sum_{k=0}^p c_k(t-1)^k$$
$$=c_pt^p+\sum_{k=0}^{p-1}c_kt^k-c_p(t-1)^p-\sum_{k=0}^{p-1}c_k(t-1)^k$$
$$=c_pt^p-c_p(t-1)^p+\sum_{k=0}^{p-1}c_kt^k-\sum_{k=0}^{p-1}c_k(t-1)^k$$
$$=c_p(t^p-(t-1)^p)+\sum_{k=0}^{p-1}c_kt^k-\sum_{k=0}^{p-1}c_k(t-1)^k$$
What I know is that when I expand $(t-1)^p$, it will give a coefficient $t^p$ that will cancel with other $t^p$ then the order of polynomial will be $p-1$, but I do not know how to develop it formally.
The answer just say

$$\nabla m_t=pc_pt^{p-1}+\sum_{k=0}^{p-2}b_kt^k$$

 A: Using the binomial formula
$$(x+y)^n = \sum _{k=1} ^n \binom n k x^n y^{n-k} ,$$
we may write
$$\nabla m_t = m_t - m_{t-1} = \sum _{k=0} ^p c_k t^k - \sum _{k=0} ^p c_k \color{blue} {(t-1)^k} = \sum _{k=0} ^p c_k t^k - \sum _{k=0} ^p c_k \color{blue} {\sum _{i=0} ^k \binom k i t^i (-1)^{k-i}} = \\
\text{[split the last sum]} = \underbrace{ \sum _{k=0} ^p c_k t^k - \sum _{k=0} ^p c_k \underbrace{ \binom k k t^k (-1)^{k-k} }_{= t^k}} _{= 0} - \sum _{k= \color{red} 1} ^p c_k \sum _{i=0} ^{\color{red} {k-1}} \binom k i t^i (-1)^{k-i} = \\
\text{[let } l = k-1 \text{]} = - \sum _{l=0} ^{p-1} c_{l+1} \sum _{i=0} ^l \binom {l+1} i t^i (-1)^{l+1-i} = \\
\text{[interchange the sums]} = - \sum _{i=0} ^{p-1} t^i \sum _{l=i} ^{p-1} c_{l+1} \binom {l+1} i (-1)^{l+1-i} ,$$
and letting
$$\tilde c_i = - \sum _{l=i} ^{p-1} c_{l+1} \binom {l+1} i (-1)^{l+1-i}$$
allows us to finally write
$$\nabla m_t = \sum _{i=0} ^{p-1} \tilde c_i t^i ,$$
showing that, indeed, $\nabla m_t$ is a polynomial of degree at most $p-1$.
Notice that if $m_t$ is a polynomial of degree exactly $p$, then $c_p \ne 0$. On the other hand,
$$\tilde c _{p-1} = - \sum _{l = p-1} ^{p-1} c_{l+1} \binom {l+1} {p-1} (-1)^{l+1-(p-1)} = - c_p \binom p {p-1} (-1)^1 = p c_p ,$$
so if $c_p \ne 0$, it follows that $\tilde c _{p-1} \ne 0$, so $\nabla m_t$ has degree exactly $p-1$.
It is easy to reason recursively, now, in order to deduce that $\nabla ^2 m_t$ is a polynomial of order $p-2$, and, in general, $\nabla ^k m_t$ is a polynomial of order $p-k$ for $0 \le k \le p$. In particular, $\nabla ^p m_t$ is a polynomial of order $p-p = 0$, so it is a constant, therefore $\nabla ^{p+1} m_t = 0$.
