I want to prove that the derivative of the $n$th order Taylor polynomial is the $n-1$th order Taylor polynomial of the derivative. More specifically:
Suppose $f: \mathbb{R} \to \mathbb{R}$ is $n$ times differentiable about $x_0$ and $T_n(x)$ a polynomail of degree at most $n$. Then if $$ \lim_{x\to x_0}\frac{f(x) - T_n(x)}{(x-x_0)^n} = 0 $$ then $$ \lim_{x\to x_0}\frac{f'(x) - T'_n(x)}{(x-x_0)^{n-1}} = 0 $$
$\textbf{I am looking for a proof of this fact that is as direct as possible}$. The fact looks like a simple corollary of l'hopitals rule. However I don't think this is legitimate since l'hopitals rule assumes that the limit $\frac{f'}{g'}$ exists, and this is what we are trying to prove.
I think this property should be true by the following argument: The hypothesis implies $T_n(x)$ is the $n$th order Taylor polynomial of $f$ about $x_0$. Hence $$ T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k. $$ This implies $$T'_n(x) = \sum_{k=0}^{n-1} \frac{f^{(k+1)}(x_0)}{k!}(x-x_0)^k = \sum_{k=0}^{n-1} \frac{g^{(k)}(x_0)}{k!}(x-x_0)^k$$
where $g(x) = f'(x)$. Hence $T'_n(x)$ is the $n-1$th order Taylor polynomial of $g(x) = f'(x)$. Again it is well known that this implies $$\lim_{x\to x_0}\frac{f'(x) - T'_n(x)}{(x-x_0)^{n-1}} = 0$$
This is not true as stated. Take $n=1$. Then if $T_1=a+b(x-x_0)$, the first condition just means that $f(x_0)=a$ and $f'(x_0)=b$. From that, you want to conclude that $\lim_{x\to x_0} f'(x)=b$, i. e. that $f'$ is continuous at $x_0$. But there are differentiable functions with discontinuous derivatives, e. g. $x^2\sin(1/x)$, extended to be $0$ at $x=0$.
If $n>1$, then your proof is fine. The catch is that to apply Taylor's theorem to $g$, you need $n-1\geq 1.$ (or you can extend it to polynomials of degree 0 is you define a 0 times differentiable function to be a continuous function, but then it is not true that a derivative of an $n$ times differentiable function is $n-1$ times differentiable).