Is there a simple proof that a Taylor polynomial of a polynomial equals itself?

Question

Say you have a polynomial $p(x)$ of degree $n$ written in standard form and you want to shift it so it is centered at $c \neq 0$. The Taylor polynomial of $p(x)$ of degree $n$ centered at $c$ gives the recentered polynomial. Because the sums are finite, I was thinking maybe there is a way to prove this that doesn't require pulling out Taylor's Theorem, i.e., the Lagrange form of the error. Is there such a proof?

Answer 1

$$p(x)=\sum_ka_k\frac{x^k}{k!}$$ $$p'(x)=\sum_ka_k\frac{x^{k-1}}{(k-1)!}=\sum_ka_{k+1}\frac{x^k}{k!}$$ $$p^{(m)}(x)=\sum_ka_{k+m}\frac{x^k}{k!}$$ $$T_c(x)=\sum_mp^{(m)}(c)\frac{(x-c)^m}{m!}$$ $$=\sum_m\sum_ka_{k+m}\frac{c^k}{k!}\frac{(x-c)^m}{m!}$$ $$=\sum_{l,m,k\\l=k+m}a_l\frac{c^k}{k!}\frac{(x-c)^m}{m!}$$ $$=\sum_la_l\sum_{m,k\\k+m=l}\frac{c^k}{k!}\frac{(x-c)^m}{m!}$$ $$=\sum_la_l\frac{(c+(x-c))^l}{l!}$$ $$=\sum_la_l\frac{x^l}{l!}=p(x)$$