Evaluate for $n \in \mathbb N$ $\int_{C(i,2)} \frac{e^z}{(z-1)^n}$

Question

Evaluate for $n \in \mathbb N$ $\int_{C(i,2)} \frac{e^z}{(z-1)^n}dz$

My Understanding

$f(z)=\sum_{n=0} a_n(z-\xi)^n$

$\frac{f^n(\xi)}{(n)!} = a_n = \frac{1}{2\pi i } \int_C \frac{f(z)}{(z-\xi)^{n+1}}dz$

$\frac{f^{(n-1)}(\xi)}{(n-1)!} = a_{n-1} = \frac{1}{2\pi i } \int_C \frac{f(z)}{(z-\xi)^{n}}dz$

If $\xi = 1$

$\int_C \frac{f(z)}{(z-\xi)^{n}}dz$ = $\frac{2\pi i f^{(n-1)}(1)}{(n-1)!}$

Here $f(z) = e^z$

So do I just get the $(n-1)^{th}$ derivative of $e^z$ and sub in $1$?

Is this correct or is there an easier method of doing this question?