The integral cannot be evaluated in the usual sense when $r=1$ since there is a pole of order $4$ on the contour (at the origin). Remember that the $p$-test says that only poles of order $<1$ are integrable.
The usual method to "assign a value" to divergent integrals like this is the Cauchy principal value method, which in this case would look like
$$
\begin{align*}
&\text{PV} \oint_\gamma \frac{e^z-1}{z^5}\,dz \\
&\qquad = -5 \cdot \lim_{\epsilon \to 0^+} \left[ \int_0^{\pi-\epsilon}\frac{\exp\left(1+e^{it}\right)-1}{\left(1+e^{it}\right)^5}\,ie^{it}\,dt + \int_{\pi+\epsilon}^{2\pi} \frac{\exp\left(1+e^{it}\right)-1}{\left(1+e^{it}\right)^5}\,ie^{it}\,dt \right].
\end{align*}
$$
Here I've parameterized the circle by $z = 1+e^{it}$ but removed the arc of the circle of length $2\epsilon$ which passes through the pole at $z=0$, as in the picture below.
However, the even order of the pole presents an issue. As $\epsilon \to 0$ we'll be approaching the pole along paths tangent to the imaginary axis, and along this axis the real part of the integrand is even:
$$
\operatorname{Re}\left[\frac{e^{iy}-1}{(iy)^5}\right] = \frac{\sin y}{y^5}.
$$
Here's a plot of this:
So, since
$$
\lim_{y \to 0} \frac{\sin y}{y^5} = \infty,
$$
no cancellation will occur when calculating the principal value and the result as $\epsilon \to 0$ will be $\infty$.
