Supose we have $f:I \to \mathbb{R}$ a $C^{\infty}$ function and let $I=(a- \delta, a+\delta)$
$\forall x \in I$ we can write the function as the power series $f(x) = \sum_{n = 0}^{\infty} a_n (x - x_0)^n$
I need to show that the coefficients are exactly the Taylor coefficients: $a_n = \frac{f^{(n)} (x_0)}{n!}$
I know the standard approach would be to argue we can derive each term but since we don't know if the series of the derivatives converge uniformly, we would have to show that every power series converge uniformly in a compact inside its convergence interval and also show the series of derivatives has the same convergence radius of the original series.
I know I am being overly cautious here, but I would like to know if there is another way. For example, using induction to show what the formula of the n-th derivative of the series would be... Any comment would help!
It may be the case that your series converges only for $x=x_0$; in that case, you cannot say anything about the coefficients.
If you know that the series converges for some $x_1>x_0$ (similar argument if $x_1<x_0$) then you can easily show that the series converges uniformly on $(x_0-c,x_0+c)$, where $c=x_1-x_0$.
Now the ratio test will show that for a power series the interval of convergence for the derivatives is the same as for the original series. It follows that the series of derivatives also converges uniformly on $(x_0-c,x_0+c)$. As a consequence you can now happily differentiate term by term to obtain your expression for $a_n$.