A nice way to remember a formula is to connect it to something you already know. For example, to remember the Cauchy integral formula, I remember that $f(z\_0)=\frac{1}{2\pi i}\int\_\Gamma\frac{f(z)}{z-z\_0}\,dz$ because if you write out the Laurent series for $\frac{f(z)}{z-z\_0}$ you're likely to get something like:

$$ \frac{f(z)}{z-z\_0}=\frac{a\_{-1}}{z-z\_0}+a\_0 +a\_1(z-z\_0)+\dots \,\,\,\,(1)$$

If you take the integral of the right hand side of (1), you're going to be left with $a\_{-1} 2\pi i$, but we were really looking for the behavior of $f(z)$ at $z\_0$, and since we used an integral to find the behavior we got $2\pi i$ as an artifact, so we're just going to divide our answer by $2\pi i$ and be left with $f(z\_0)$ a.k.a $Res(f;z\_o).$

My questions: 

* is there a similar way to think about $f^{(n)}(z\_0)=\frac{n!}{2\pi i}\int\_\Gamma\frac{f(z)}{(z-z\_0)^{n+1}}\,dz$? I've tried writing out a Laurent series for $\frac{f(z)}{(z-z\_0)^{n+1}}\,dz$, but it didn't get me anywhere.
* if $f(z)=\frac{g(z)}{(z-z\_0)^m}$, is $g(z\_0)=a\_{-1}$ ? (where $a\_{-1}$ is from its Laurent series)

Obviously there is a strong connection between the singularities of a function, its Laurent series, residue theory and the Cauchy integral formulas. After a semester of introductory complex analysis, I'm struggling to place everything in a coherent narrative. Any help would be greatly appreciated.

**Edit**: Based on the answers I've formed a new narrative to remember the formulas. I thought I'd share it in case someone finds it helpful.

Let's say we have a function:

$$ \frac{ f(z) }{ (z-z\_0)^{n+1} } $$

We want to study the behavior of $f^{(n)}(z)$ at $z=z\_0$. We know that a Taylor coefficient is on the form $a\_n=\frac{f^{(n)}(z\_0)}{n!}$, so we can find $f^{(n)}(z\_0)$ by manipulating the Taylor series.

So expand to Taylor form:

$$ \frac{ f(z) }{ (z-z\_0)^{n+1} }=\frac{c\_0}{(z-z\_0)^{n+1}}+\frac{c\_{1}}{(z-z\_0)^{n}} +\dots+\frac{c\_{k}}{(z-z\_0)^{1}}+c\_{k+1}+c\_{k+2}(z-z\_0)+\dots $$ 

We see that we can recover $c\_k=\frac{f^{(k)}(z\_0)}{n!}$ by integrating the equation above. But since we used an integral to isolate $c\_k$, we get $2\pi i$ as an artifact, and since we used the Taylor series we also got $\frac{1}{n!}$ as an artifact.

We can think of the forumla $f^{(n)}(z\_0)=\frac{n!}{2\pi i}\int\_\Gamma\frac{f(z)}{(z-z\_0)^{n+1}}$ a formula for recovering $f^{(n)}(z\_0)$ from a Taylor series: Use $\int\_\Gamma\frac{f(z)}{(z-z\_0)^{(n+1)}}$ to isolate the Taylor coefficient and remove the extraneous information by multiplying by $\frac{n!}{2\pi i}$.

When $n=0$ it's just the regular Cauchy integral formula, of course; $\frac{0!}{2\pi i}=\frac{1}{2\pi i}$.