We know that if $f(z)$ is analytic at $a \in \mathbb{C}$, then by Taylor,
$$f(z) =\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(z-a)^n$$
By Cauchy, we also know that
$$f^{(n)}(a) = \frac{n!}{2 \pi i } \int_C \frac{f(s)}{(s-a)^{n+1}} \, ds$$
Combining these two, I'm imagining that it would then follow that
$$f(z) =\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(s-a)^n = \sum_{n=0}^\infty \frac{\frac{n!}{2 \pi i } \int_C \frac{f(s)}{(s-a)^{n+1}} \hspace{0.1cm} ds}{n!}(z-a)^n $$
$$ = \sum_{n=0}^\infty \frac{1}{2 \pi i } \left[\int_C \frac{f(s)}{(s-a)^{n+1}} \hspace{0.1cm} ds \right] (z-a)^n $$
I've become very familiar with Taylor series and Cauchy's integral formula as separate understandings, but I've never seen anyone combine both of the results. Wouldn't this be a representation of the Taylor formula that complex analysis textbooks should include as a direct corollary of both Taylor and Cauchy Integral Formula? Why haven't I seen these two popular tools combined in this manner? Perhaps I have made an error?