Let $\displaystyle \text{deg}(p)=n$ fix $\displaystyle x_0\in\mathbb{R}$ (where $\displaystyle \mathbb{R}_n[x]$ is just the set of all $\displaystyle n^{\text{th}}$-degree polynomials of degree $\displaystyle n$) and let $\displaystyle D_{x_0}^j:\mathbb{R}_n[x]\to\mathbb{R}_n[x],\quad j\in\{0,\cdots,n\}$ be the mapping $\displaystyle \left(D_{x_0}^j(p)\right)(x)=(x-x_0)^j p^{(j)}(x_0)$. This mapping is linear since

$\displaystyle \begin{aligned}\left(D_{x_0}^j(\alpha p+\beta q)\right)(x) &= (x-x_0)^j\left(\alpha p+\beta q)^{(j)}(x)\\ &= (x-x_0)^j (\alpha p^{(j)}+\beta q^{(j)})(x_0)\\ &= (x-x_0)^j\left(\alpha p^{(j)}(x_0)+\beta q^{(j)}(x_0)\right)\\ &= \alpha (x-x_0)^j p^{(j)}(x_0)+\beta (x-x_0)^j q^{(j)}(x_0)\\ &= \alpha\left(D_{x_0}^j(p)\right)(x)+\beta\left(D_{x _0}^j(q)\right)(x)\end{aligned}$

and thus the function

$\displaystyle T_n(p):\mathbb{R}_n[x]\to\mathbb{R}_n[x]$

where

$\displaystyle \displaystyle \left(T_n(p)\right)(x)=\sum_{j=0}^{n}\frac{1}{j!}D _{x_0}^j(p)$

is linear. But, since $\displaystyle \{1,\cdots,x^n\}$ is a basis for $\displaystyle \mathbb{R}_n[x]$; to prove that $\displaystyle T_n(p)=p$ for all $\displaystyle p\in\mathbb{R}_n[x]$ it suffices to check it for $\displaystyle p(x)=x^j,\text{ }j=0,\cdots,n$. To do this we merely note that

$\displaystyle \displaystyle \begin{aligned}T_n(x^m) &= \sum_{j=0}^{n}\frac{1}{j!}D^j_{x_0}(x^m)\\ &= \sum_{j=0}^{m}\frac{m(m-1)\cdots(m-j+1)}{j!}(x-x_0)^j x_0^{m-j}\\ &= \sum_{j=0}^{m}{m\choose j}(x-x_0)^j x_0^{m-j}\\ &= (x-x_0+x_0)^m\\ &=x^m\end{aligned}$

where we've used the fact that after the $\displaystyle m^{\text{th}}$ term the terms vanish (that's how we can account for the change of the upper index) and at the last step we employed the Binomial theorem. The conclusion follows by noticing that $\displaystyle T_n(p)$ is the $\displaystyle n^{\text{th}}$ degree Taylor polynomial at $\displaystyle x_0$, where of course we realize that the remainder term is zero since the $\displaystyle n+1^{\text{th}}$ derivative of $\displaystyle p$ is identically zero.