Proof of the theorem about term-wise differentiation of power series

I have some doubts concerning the proof of the "term-wise differentiation of power series" theorem. Below, I first included 3 theorems that are used in the proof; then, I included the whole proof and state the related theorems; finally, I included my specific doubts about it.

The theorems used in the proof are basically the Squeeze Theorem and the facts that the first and second derivatives of a power series have the same radius of convergence as the original power series. They are stated as follows:

Theorem 1: If the radius of convergence of the power series $\sum_{n=0}^{+\infty} c_nx^n$ is $R>0$, then R is also the radius of convergence of the series $\sum_{n=1}^{+\infty} nc_nx^{n-1}.$

Theorem 2: If the radius of convergence of the power series $\sum_{n=0}^{+\infty} c_nx^n$ is $R>0$, then R is also the radius of convergence of the series $\sum_{n=2}^{+\infty} n(n-1)c_nx^{n-2}.$

Theorem 3 (also known as Squeeze Theorem): Assume $f$, $g$ and $h$ are functions defined in an open interval $I$ containing $a$, except, possibly, at $a$ itself, and that $f(x) \leq g(x) \leq h(x)$ for all $x$ in $I$ such that $x\neq a$. If both $\lim_{x\to a}f(x)$ and $\lim_{x\to a}h(x)$ exist and are equal to $L$, then $\lim_{x\to a}g(x)$ also exists and is equal to $L$.

The theorem statement and its proof, about which I have some questions, are below:

Theorem: Let $\sum_{n=0}^{+\infty} c_nx^n$ be a power series whose radius of convergence is $R>0$. Then, if $f$ is the function defined by

$$f(x) = \sum_{n=0}^{+\infty} c_nx^n \ \ \ \ (1)$$

$f'(x)$ exists for all $x$ in the open interval $(-R,R)$, and it is given by

$$f(x) = \sum_{n=1}^{+\infty} n c_nx^{n-1}$$

PROOF: Let $x$ and $a$ be two distinct numbers in the open interval $(-R,R)$. The Taylor formula, with $n = 1$, is

$$f(x) = f(a) + \dfrac{f'(a)}{1!}(x-a) + \dfrac{f''(\xi)}{2!}(x-a)^2$$

Using this formula with $f(x) = x^n$, we have, for every positive integer $n$, $$x^n=a^n+na^{n-1}(x-a)+\frac{1}{2}n(n-1)(\xi_n)^{n-2}(x-a)^2 \ \ \ \ (2)$$ where $\xi_n$ is between $a$ and $x$, for every positive integer $n$. From (1) we have \begin{align} f(x)-f(a) &= \sum_{n=0}^{+\infty} c_nx^n - \sum_{n=0}^{+\infty} c_na^n\\ &= c_0 + \sum_{n=1}^{+\infty} c_nx^n - c_0 - \sum_{n=1}^{+\infty} c_na^n\\ &= \sum_{n=1}^{+\infty} c_n(x^n - a^n) \end{align} Dividing by $x-a$ (because $x\neq a$) and using (2), we have, from the above equation $$\dfrac{f(x)-f(a)}{x-a} = \dfrac{1}{x-a} \sum_{n=1}^{+\infty} c_n[na^{n-1}(x-a)+\frac{1}{2}n(n-1)(\xi_n)^{n-2}(x-a)^2]$$ Thus, $$\dfrac{f(x)-f(a)}{x-a} = \sum_{n=1}^{+\infty} nc_na^{n-1}+\frac{1}{2}(x-a)\sum_{n=2}^{+\infty}n(n-1)c_n(\xi_n)^{n-2} \ \ \ \ (3)$$

Since $a$ is in $(-R,R)$, it follows from Theorem 1 that $\sum_{n=1}^{+\infty}nc_na^{n-1}$ is absolutely convergent.

Since both $a$ and $x$ are in $(-R,R)$, there exists a number $K > 0$ such that $|a|<K<R$ and $|x|<K<R$. It follows from Theorem 2 that

$$\sum_{n=2}^{+\infty}n(n-1)c_nK^{n-2}$$

is absolutely convergent. Then, since

$$|n(n-1)c_n(\xi_n)^{n-2}| < |n(n-1)c_nK^{n-2}| \ \ \ \ (4)$$

for every $\xi_n$, we can conclude, from the comparison test, that $$\sum_{n=2}^{+\infty}n(n-1)c_n(\xi_n)^{n-2}$$ is absolutely convergent.

It follows from (3) that

$$\left|\dfrac{f(x)-f(a)}{x-a} - \sum_{n=1}^{+\infty} nc_na^{n-1}\right| = \left|\frac{1}{2}(x-a)\sum_{n=2}^{+\infty}n(n-1)c_n(\xi_n)^{n-2}\right| \ \ \ \ (5)$$

However, we know that, if $\sum_{n=1}^{+\infty}u_n$ is absolutely convergent, then

$$\left|\sum_{n=1}^{+\infty}u_n\right| \leq \sum_{n=1}^{+\infty}|u_n|$$

Applying this to the right side of (5), we obtain: $$\left|\dfrac{f(x)-f(a)}{x-a} - \sum_{n=1}^{+\infty} nc_na^{n-1}\right| \leq \frac{1}{2}|x-a|\sum_{n=2}^{+\infty}n(n-1)|c_n||\xi_n|^{n-2} \ \ \ \ (6)$$ From (4) and (6), we obtain: $$\left|\dfrac{f(x)-f(a)}{x-a} - \sum_{n=1}^{+\infty} nc_na^{n-1}\right| \leq \frac{1}{2}|x-a|\sum_{n=2}^{+\infty}n(n-1)|c_n|K^{n-2} \ \ \ \ (7)$$ where $0<K<R$. Since the series on the right side of (7) is absolutely convergent, the limit on the right side, as $x$ approaches $a$, is zero. Thus, from (17) and from Theorem 3 (Squeeze Theorem), it follows that $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\sum_{n=1}^{+\infty}nc_na^{n-1}$$ which is equivalent to $$f'(a)=\sum_{n=1}^{+\infty}nc_na^{n-1}$$ and, since $a$ can be any number in the open interval $(-R,R)$, the theorem is proven.

My 2 doubts are:

1. Why is the step that transforms (6) into (7) necessary? Assuming that the goal is to have the right side of the inequality to be convergent, then, since $\sum_{n=2}^{+\infty}n(n-1)c_n(\xi_n)^{n-2}$ is known to be absolutely convergent, isn't this goal already achieved by the summation on the right side of (6)?
2. How is the last limit obtained from (7) and the Squeeze Theorem? It seems that, from (7) and the Squeeze Theorem, it follows that $\lim_{x\to a} \left[ \dfrac{f(x)-f(a)}{x-a} - \sum_{n=1}^{+\infty} nc_na^{n-1}\right] =0$. So, I think that, since $f(x)$ is differentiable at $x = a$, then I know that $lim_{x\to a} [f(x)-f(a)]/(x-a)$ exists; so I can simply distribute the limit over the sum and arrive at the result. Is this correct?
-

For question 2, the application of the Squeeze Theorem that they have in mind is likely the following. If $C$ is the sum of the series on the right-hand side of (7), we can write the inequality (7) as
$$\sum_{n=1}^\infty n c_n a^{n-1} - \frac{1}{2} |x-a| C \leq \frac{f(x) - f(a)}{x - a} \leq \sum_{n=1}^\infty n c_n a^{n-1} + \frac{1}{2} |x-a|C.$$
Letting $x \to a$, both the left and right sides of this double-inequality tend to $\sum_{n=1}^\infty n c_n a^{n-1}$, so the limit of the central expression must also exist and have the same value.