I'm trying to show that $$ (-1)^{n+1} \frac{d^n}{dt^n}(1 - e^{-t})^\alpha = -(x \frac{d}{dx})^n(1-x)^\alpha$$
where $x = e^{-t}$, $\alpha > 1, n > 0$.
If I understand correctly, $$ \frac{d}{dt} = \frac{d}{d(-\ln(x))} = -x\frac{d}{dx}$$
However, it's not clear how does the second (and further) derivatives follow:
$$ \frac{d}{dt}(-x \frac{d}{dx}) = -x\frac{d}{dx}(-x \frac{d}{dx}) \stackrel{?}{=}(-1)^2 x^2 \frac{d}{dx}\frac{d}{dx}$$
Do $\frac{d}{dx}$ just multiplies like that? If so, why does this hold? That is, why isn't $\frac{d}{dx}(-x \frac{d}{dx}) = -1 \frac{d}{dx}\frac{d}{dx} -x \underbrace{\frac{d(\frac{d}{dx})}{dx}}_{=0? ~~ =d^2/dx^2?}$ as would be following the chain rule with ordinary variables (instead of $d/dx$)?
In other words, I don't understand how exactly manipulating $d/dx$ works.
Your first step is correct and basically all you need. Remember that these exponents mean "applying after another": $$\frac{\mathrm{d}^n}{\mathrm{d}t^n} f(t)= \underbrace{\frac{\mathrm{d}}{\mathrm{d}t}\left(\cdots \frac{\mathrm{d}}{\mathrm{d}t}f(t)\right)}_{\text{$n$-times}}.$$ Since you already arrived at the expression $$\frac{\mathrm{d}}{\mathrm{d}t} = -x \frac{\mathrm{d}}{\mathrm{d}x}$$ you can solve the problem now by just substituting: $$\frac{\mathrm{d}^n}{\mathrm{d}t^n} f(t)= \underbrace{\frac{\mathrm{d}}{\mathrm{d}t}\left(\cdots \frac{\mathrm{d}}{\mathrm{d}t}f(t)\right)}_{\text{$n$-times}}=\left(-x \frac{\mathrm{d}}{\mathrm{d}x}\right)^n f(t).$$ In your case: $$(-1)^{n+1} \frac{\mathrm{d}^n}{\mathrm{d}t^n} (1-e^{-t})^\alpha = (-1)^{n+1}\left(-x \frac{\mathrm{d}}{\mathrm{d}x}\right)^n (1-x)^\alpha = -\left(x \frac{\mathrm{d}}{\mathrm{d}x}\right)^n(1-x)^\alpha.$$