[proofplan] Write $p(t)$ by degree and apply the [binomial theorem](/theorems/750) to $t^k-(t-1)^k$. For $k\ge 1$, this difference has leading term $k t^{k-1}$ and only lower-degree remaining terms. Over $\mathbb{R}$ the scalar $k$ is nonzero, so the leading term of $\Delta p$ is $r a_r t^{r-1}$ when $p$ has degree $r\ge 1$. Iterating the degree drop gives a constant after $r$ differences and zero after one more difference. [/proofplan] [step:Compute the difference of a monomial] For $k\ge 1$, the binomial theorem gives \begin{align*} (t-1)^k=\sum_{\ell=0}^{k}\binom{k}{\ell}t^\ell(-1)^{k-\ell}. \end{align*} The coefficient of $t^k$ in this expansion is $1$, and the coefficient of $t^{k-1}$ is $-k$. Therefore \begin{align*} t^k-(t-1)^k=k t^{k-1}+q_k(t), \end{align*} where $q_k\in\mathbb{R}[t]$ has degree at most $k-2$ when $k\ge 2$, and $q_1=0$. Thus $\Delta(t^k)$ has degree $k-1$ for every $k\ge 1$. Also $\Delta(1)=0$. [guided] For $k\ge 1$, expand $(t-1)^k$ as \begin{align*} (t-1)^k=\sum_{\ell=0}^{k}\binom{k}{\ell}t^\ell(-1)^{k-\ell}. \end{align*} The term with $\ell=k$ is $t^k$, and the term with $\ell=k-1$ is $-k t^{k-1}$. Hence subtracting from $t^k$ cancels the degree-$k$ term and gives \begin{align*} t^k-(t-1)^k=k t^{k-1}+q_k(t), \end{align*} where $q_k\in\mathbb{R}[t]$ has degree at most $k-2$ if $k\ge 2$, while $q_1=0$. Since $k\ne 0$ in $\mathbb{R}$, the leading term $k t^{k-1}$ is nonzero. Thus $\Delta(t^k)$ has degree $k-1$ for $k\ge 1$, and $\Delta(1)=0$. [/guided] [/step] [step:Apply the monomial computation to a polynomial of exact degree] Assume first that $r\ge 1$, and write \begin{align*} p(t)=a_r t^r+\sum_{k=0}^{r-1}a_k t^k \end{align*} with $a_r\ne 0$. By linearity of $\Delta$, \begin{align*} (\Delta p)(t)=a_r\Delta(t^r)+\sum_{k=0}^{r-1}a_k\Delta(t^k). \end{align*} The first term has leading term $r a_r t^{r-1}$. For each $k\le r-1$, the polynomial $\Delta(t^k)$ has degree at most $k-1\le r-2$, with the constant case giving zero. Hence no lower-degree summand can cancel the coefficient of $t^{r-1}$. Because $r a_r\ne 0$ in $\mathbb{R}$, the polynomial $\Delta p$ has degree $r-1$. [/step] [step:Treat constants and iterate] If $r=0$, then $p(t)=a_0$ is constant, so \begin{align*} (\Delta p)(t)=a_0-a_0=0. \end{align*} For $r\ge 1$, the previous step shows that each application of $\Delta$ lowers the degree by exactly one as long as the current polynomial has positive degree. Thus $\Delta^m p$ has degree $r-m$ for $0\le m\le r$. In particular, $\Delta^r p$ is constant, and applying $\Delta$ once more gives \begin{align*} \Delta^{r+1}p=0. \end{align*} This proves both the one-step degree statement and the eventual annihilation of polynomial trends. [/step]