$p(x)$ divided by $x-c$ has remainder $p(c)$? [Polynomial Remainder Theorem] This is from Pinter, A Book of Abstract Algebra, p.265. 
Given $p(x) \in F[x]$ where $F$ is a field, I would like to show that $p(x)$ divided by $x-c$ has remainder $p(c)$.
This is easy if $c$ is a root of $p$, but I don't see how to prove it if $c$ is not a root.

Suppose that
$$
p(x)=\sum_{k=0}^na_kx^k
$$
Then
$$
p(x)-p(c)=\sum_{k=0}^na_k(x^k-c^k)
$$
For each $k$, we have that
$$
\frac{x^k-c^k}{x-c}=x^{k-1}+x^{k-2}c+x^{k-3}c^2+\dots+xc^{k-2}+c^{k-1}
$$
Thus,
$$
\begin{align}
\frac{p(x)-p(c)}{x-c}
&=\sum_{k=0}^na_k(x^{k-1}+x^{k-2}c+x^{k-3}c^2+\dots+xc^{k-2}+c^{k-1})\\
&=q(x)
\end{align}
$$
Therefore, $p(x)=q(x)(x-c)+p(c)$, which is another way of writing $p(x)$ divided by $x-c$ leaves a remainder of $p(c)$ since the degree of $p(c)$, $0$, is less than the degree of $x-c$, $1$.
