Generally speaking, you use Cauchy's theorem for the following integral:

$$\frac1{i 2 \pi} \oint_C dz \, z^{(1+n) \alpha-1} \, e^z $$

Yes, you provide a keyhole contour $C$ about the negative real axis and evaluate the integral along each piece of the contour and set the contour integral to zero (or, the sum of the residues of the poles if there are any; there are none here).

For a picture of the contour $C$, see this answer. The integral vanishes along the outer circular arcs only for certain values of $n$. To find those values, consider the integral about the arc in the upper half-plane, which upon parametrization by $z=R e^{i \theta}$, where $\theta \in [\pi/2,\pi)$. Thus, the integral there is

$$\frac1{2 \pi} \int_{\pi/2}^{\pi} d\theta \, e^{i \theta} \, R^{(1+n) \alpha} e^{i (1+n) \alpha \theta} e^{R\cos{\theta}} e^{i R \sin{\theta}}$$

We now consider the upper bound of the magnitude of this integral (as it is not really in a good form for exact evaluation). Note that, in general,

$$\left | \int_{\gamma} dz \, f(z) \right | \le \int_{\gamma} |dz|\, |f(z)|$$

Let's apply this principle to the integral over $\theta$, but let's also subtract $\pi/2$ from $\theta$ for later convenience. The magnitude of this integral is now bounded by:

$$\frac1{2 \pi} R^{(1+n)\alpha} \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}}$$

Convenienetly, we may use the inequality $\sin{\theta} \ge 2 \theta/\pi$ for $\theta \in [0,\pi/2]$. Thus, we get a final upper bound for the magnitude of the integral:

$$\frac1{2 \pi} R^{(1+n)\alpha} \int_0^{\pi/2} d\theta \, e^{-R 2 \theta/\pi} \le \frac14 R^{(1+n)\alpha-1}$$

As $R \to \infty$, we hope that the integral vanishes. For this to happen, $(1+n)\alpha -1 \lt 0$, or $n+1 \lt 1/\alpha$.

The same analysis applies to the integral about the arc in the lower half-plane.

You can also apply a similar analysis to the integral about the inner circle, except now we parametrize as $z=\epsilon e^{i \phi}$, where $\phi \in (-\pi,\pi)$. Taking the limit as $\epsilon \to 0$, we find that the integral vanishes as $\epsilon^{(1+n) \alpha}$.

Thus you are left with three pieces of $C$ along which the integral does not vanish:

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} dz\, z^{(1+n) \alpha -1} \, e^z + \frac{e^{i (1+n) \alpha \pi}}{i 2 \pi} \int_{\infty}^0 dx \, x^{(1+n) \alpha-1} \, e^{-x} \\+ \frac{e^{-i (1+n) \alpha \pi}}{i 2 \pi} \int_0^{\infty} dx \, x^{(1+n) \alpha-1} \, e^{-x}$$

The first integral is the original ILT. The second is just above the real axis, where we parametrize by $z=e^{i \pi} x$. The third is just below, where $z=e^{-i \pi} x$. Note that we are respecting the branch cut along the negative real axis.

Since there are no poles inside $C$, the contour integral is zero by Cauchy's theorem. Thus we have

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} dz\, z^{(1+n) \alpha -1} \, e^z = \frac1{\pi} \sin{\pi (1+n)\alpha} \int_0^{\infty} dx \, x^{(1+n) \alpha-1} \, e^{-x}$$

Use the fact that

$$\int_0^{\infty} dx \, x^{y-1} e^{-x} = \Gamma(y)$$

to get that

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} dz\, z^{(1+n) \alpha -1} \, e^z = \frac1{\pi} \sin{ [\pi (1+n)\alpha}] \, \Gamma((1+n) \alpha)$$

Use the reflection formula

$$\Gamma(y) \Gamma(1-y) = \frac{\pi}{\sin{\pi y}}$$

and I get, finally,

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} dz\, z^{(1+n) \alpha -1} \, e^z = \frac1{\Gamma(1-(n+1) \alpha)}$$