We have:
$$f(a) = \frac{1}{2\pi i}\int_{D(z_0,r)}\frac{f(z)}{z-a}dz$$
$$f'(a) = \frac{-1}{2\pi i}\int_{D(z_0,r)}\frac{f(z)}{(z-a)^2}dz$$ $$f^{(2)}(a) = \frac{2}{2\pi i}\int_{D(z_0,r)}\frac{f(z)}{(z-a)^3}dz$$ $$f^{(3)}(a) = \frac{-3\cdot 2}{2\pi i}\int_{D(z_0,r)}\frac{f(z)}{(z-a)^3}dz$$
$$\vdots$$
$$f^{(n)}(a) = (-1)^{n}\frac{n!}{2\pi i}\int_{D(z_0,r)}\frac{f(z)}{(z-a)^{n+1}}dz$$
Why is the part $(-1)^n$ left out if we look at books or lectures. We've done it without that part, but why?
Because it shouldn't be there.
If you differentiate $\frac1{z-a}$ with respect to $a$, what you get is $\frac1{(z-a)^2}$, rather than $-\frac1{(z-a)^2}$. So,$$f'(a)=\frac1{2\pi i}\int_{D(z_0,r)}\frac1{(z-a)^2}\,\mathrm dz.$$And so on…