I don't see why Cauchy's theorem or the residue formula are needed here.
First of all, suppose
$n \ne -1; \tag 1$
then the function
$F_n(z) = \dfrac{z^{n + 1}}{n + 1} \tag 2$
is everywhere a primitive for $z^n$:
$F'_n(z) = \left ( \dfrac{z^{n + 1}}{n + 1} \right )' = z^n, \; \forall z \in \Bbb C; \tag 3$
it is then elementary that, for $z_0, z \in \Bbb C$,
$\dfrac{z^{n + 1}}{n + 1} - \dfrac{z_0^{n + 1}}{n + 1} = \displaystyle \int_{z_0}^z F'_n(w) \; dw = \int_{z_0}^z w^n \; dw; \tag 4$
for any closed path $\gamma(t)$ in $\Bbb C$,
$\gamma:[a, b] \to \Bbb C, \; \gamma(a) = \gamma(b) = z_0, \tag 5$
since $\gamma(t)$ both starts and ends at $z_0$, (4) becomes
$\displaystyle \int_\gamma w^n \; dw = \dfrac{z_0^{n + 1}}{n + 1} - \dfrac{z_0^{n + 1}}{n + 1} = 0; \tag 6$
taking
$w = \gamma(t) = c + r e^{it}, \; t \in [0, 2\pi], \tag 7$
we recover the specific case
$\displaystyle \int_\gamma (c + re^{it})^n rie^{it} \; dt = \int_0^{2\pi} (c + re^{it})^n rie^{it} \; dt = 0, n \ne -1; \tag 8$
when $n = -1$, the expression (2) cannot yield a primitive for $f(z) = z^{-1}$, since then $n + 1 = 0$; in the light of the hypothesis that $r < c$, however, we may set
$F_{-1}(z) = \ln z \tag 9$
in an open disk $D(c, c + \epsilon)$, $\epsilon < c - r$; then
$F'_{-1}(z) = \dfrac{1}{z} \tag{10}$
in $D(c, c + \epsilon)$, so essentially the same argument as in (4), (6) applies; for
$z_0, z \in D(c, c + \epsilon)$,
$\ln z - \ln z_0 = \displaystyle \int_{z_0}^z F'_{-1}(w) \; dw = \int_{z_0}^z \dfrac{dw}{w}; \tag{11}$
$\displaystyle \int_\gamma \dfrac{dw}{w} = \ln z_0 - \ln z_0 = 0, \tag{12}$
$\gamma(t)$ a closed path in $D(c, c + \epsilon)$.
Best Answer
$\newcommand{\d}{\mathrm{d}}$Unfortunately, the majority of complex integrals (or just the majority of all integrals) that can be easily written down can't be easily found if you only use techniques of Riemann integration. In general, it is good advice to completely avoid substitutions $z=\gamma(t)$! Complex analysis has produced many beautiful theorems involving integration of complex contours.
The one I will present to you now is the Cauchy Integral Formula:
If $f$ is a holomorphic function in some neighbourhood of $s\in\Bbb C$, and $\gamma$ is a closed contour that is contained in this neighbourhood which also winds counterclockwise - once - about the point $s$.
Here, $s=-1$ and the circle of radius $2$ - I assume a counterclockwise winding - is a contour of the correct type as it encloses $-1$. $f(z)=e^z$ is of course holomorphic in this neighbourhood of $-1$. It follows that: $$\frac{\d^2}{\d z^2}[e^z]\Big |_{z=-1}=\frac{2!}{2\pi i}\oint_{\gamma}\frac{e^z}{(z+1)^3}\,\d z$$But that just means to say your integral equals $\frac{\pi i}{e}$.