Show that:
$$\sum^{\infty}_{n=k}n(n-1)...(n-k+1)x^{n-k}=\frac{k!}{(1-x)^{k+1}}~,$$ where $\vert x\vert < 1 $.
My proof:
EDITED
Let's consider : $S_n:=\sum^{\infty}_{n=0}x^{n}$, where $ \vert x \vert < 1$. We know that it is simply a geometric series and hence pointwise convergent on $ \vert x \vert < 1$. Further, the partial sums are infinitely differentiable as they are polynomials.
We now construct a power series $lim_{m \to \infty}\sum^{m}_{n=0}\frac{d^k}{dx^k}x^{n}$ which consists of the $k$-th derivative of the partial sums of $S_n$. It holds that: $lim_{m \to \infty}\sum^{m}_{n=0}\frac{d^k}{dx^k}x^{n}=lim_{m \to \infty}\sum^{m}_{n=k}n(n-1)...(n-k+1)x^{n-k}$, as the first $k$-many summands are $0$. To check for uniform convergence I will calculate the limes superior of the following sequence of ratios, where $n \geq k$:
$$lim ~sup_{n\to \infty} \bigg| \frac{(n+1)n(n-1)...(n-k+2)x^{n+1-k}}{n(n-1)...(n-k+1)x^{n-k}}\bigg| =lim ~sup_{n\to \infty}\bigg|\frac{n+ 1}{n-k +1}x\bigg|=x. $$ The radius of convergence $R$ is: $\vert x\vert < 1=:R$, with $R \in \mathbb{R}$.
So I know that the power series $lim_{m \to \infty}\sum^{m}_{n=0}\frac{d^k}{dx^k}x^{n}$ converges pointwise on $]-R, +R[$ and that it converges uniformly on any compact set $[\alpha, \beta] \subseteq ~]-R, +R[$, with $\alpha, \beta \in \mathbb{R}$.
Due to uniform convergence, we are allowed to swap the derivative operator and the limit. $\sum^{\infty}_{n=0}\frac{d^k}{dx^k}x^{n}=\frac{d^k}{dx^k}\sum^{\infty}_{n=0}x^{n}=\frac{d^k}{dx^k}~lim_{n \to \infty}\frac{1-x^{n}}{1-x}= \frac{d^k}{dx^k}~\frac{1}{1-x}$. Via induction we see: $\frac{d^k}{dx^k}~\frac{1}{1-x}= \frac{k!}{(1-x)^{k+1}}$. q.e.d.
So the limit of the initial power series $\sum^{\infty}_{n=k}n(n-1)...(n-k+1)x^{n-k}=\frac{k!}{(1-x)^{k+1}}~,$ does not exist uniformly for all $\vert x \vert < 1$ but only on compact sets, like $[\alpha, \beta] \subseteq ~]-R, +R[$.
Is this proven rigorously or am I missing something?
Your proof has the right steps. But
Absolutely not. $\sum^{\infty}_{n=0} 2^n x^n$ has radius of convergence $1/2$. $\sum^{\infty}_{n=0} n! x^n$ has radius of convergence $0$.
Also, $\sum^{\infty}_{n=0} x^n $ is not uniformly convergent for $|x|<1$.
To see this, just note that any $N$th partial sum extends to a continuous (hence bounded) function on $|x|\le 1$, but the limiting function explodes as $x\to 1$. Thus, for each $N$, $$ \sup_{|x|<1} \left| \sum^{\infty}_{n=0} x^n - \sum^{N}_{n=0} x^n \right| = \infty $$
It is uniformly convergent on closed subsets of $|x|<1$.