Consider the contour integral $$\int_C \frac{(-x)^s}{e^x-1}\frac{dx}{x}$$ where the path of integration begins at $\infty$, moves to the left down the positive real axis, circles the origin once in the positive direction, and returns up to the positive real axis to $\infty$. The definition of $(-x)^s$ is $(-x)^s=\exp (s \log (-x))$. $\zeta (s)$ can be represented by $$\zeta (s)=\frac{\Gamma (1-s)}{2\pi i}\int_C \frac{(-x)^s}{e^x-1}\frac{dx}{x}.$$
Now in his book Riemann's Zeta Function, Edwards states
The contour integral clearly converges for all values of $s$ (because $e^x$ grows much faster than $x^s$ as $x\to \infty$).
I don't understand this explanation at all. Using this logic, we can conclude that the integral in $$\zeta (s) \Gamma (s)=\int_0^\infty \frac{x^{s}}{e^x-1}\frac{dx}{x}$$ converges for all values of $s$ "since $e^x$ grows much faster than $x^s$", which is clearly nonsense, as the integral converges only for $\Re (s)\gt 1$.
Can someone explain Edwards' reasoning?