$\int_{0}^{a}\left(\sum_{j=0}^{n-1}(-x)^j\right)^{b-1}\cdot\sum_{j=0}^{n-1}(jb+1)(-x)^j\mathrm dx=I$ Consider

$$\int_{0}^{a}\left(\sum_{j=0}^{n-1}(-x)^j\right)^{b-1}\cdot\sum_{j=0}^{n-1}(jb+1)(-x)^j\mathrm dx=I\tag1$$
  How does one show that 
  $$I=a(1-a+a^2-a^3+\cdots+a^{n-1})^{b}$$
  $(a,b)>0$

expanded
$$\int_{0}^{a}(1-x+x^2-\cdots-x^{n-1})^{b-1}(1-(b+1)x+(2b+1)x^2-\cdots[(n-1)b+1]x^{n-1})\mathrm dx=I$$

Assume $a<1$, then
$$\int_{0}^{a}(1+x)^{1-b}\cdot\sum_{j=0}^{\infty}(jb+1)(-x)^j\mathrm dx\tag2$$
But doesn't really help to prove $(1)$
 A: Define $f(x) = \sum_{j=0}^{n-1}(-x)^j$ then the integral can be written
$$I(a) = \int_{0}^{a}f(x)^{b} + x\frac{d}{dx}f(x)^{b}{\rm d}x = \int_0^a\frac{d}{dx}\left[xf(x)^b\right]{\rm d}x = af(a)^b$$
A: Winther's answer is definitely the shortest way. Here I will detail out how we get there, and show what a solution would be like if were to actually try differentiation. We first run the substitution $-x \to x$ (not necessary at all, but it makes the geometric sum more obvious for beginners in my opinion)
$$\int_{0}^{a}\left(\sum_{j=0}^{n-1}x^j\right)^{b-1}\left(\sum_{j=0}^{n-1}(jb+1)x^j\right)\mathrm dx$$
We first note that, by the geometric series formula,
$$\sum_{j=0}^{n-1}x^j = \frac{x^n -1}{x-1}\tag{1}$$
We can also split the second sum into the form
$$b\sum_{j=0}^{n-1}jx^j+\sum_{j=0}^{n-1}x^j $$
The second sum is clearly the geometric series we had before; the first can be written as
$$b\sum_{j=0}^{n-1}jx^j=bx\sum_{j=0}^{n-1}\frac{d}{dx}\left(x^j\right)=bx\frac{d}{dx}\sum_{j=0}^{n-1}x^j \tag{2}$$
We can use $(1)$ and $(2)$ we can write the original problem as
$$\int_{0}^{a}\left(\frac{x^n -1}{x-1}\right)^{b-1}\left(bx\frac{d}{dx}\left(\frac{x^n -1}{x-1}\right)+\frac{x^n -1}{x-1}\right)\mathrm dx \tag{3}$$
If we try to actually differentiate we get
$$\int_{0}^{a}\frac{\left(\frac{x^n-1}{x-1}\right)^{b-1} \left(x^n ((bn-b+1) x-b n-1)+(b-1)
   x+1\right)}{(x-1)^2}\mathrm dx$$
This could be simplified, if you had the patience of a god. I will leave that up to you though, and just point out that we can write $(3)$ in the form
$$\int_{0}^{a}f(x)^{b-1}\left(bx\frac{d}{dx}f(x)+f(x)\right)\mathrm dx$$
$$=\int_{0}^{a} \left(bx\frac{d}{dx}f(x)f(x)^{b-1}+f(x)^b\right)\mathrm dx$$
$$=\int_{0}^{a} \bigg(bxf'(x)f(x)^{b-1}+f(x)^b\bigg)\mathrm dx$$
Applying the chain rule in reverse we find that $x\frac{d}{dx}f(x)^b = f(x)^{b-1}\cdot bxf'(x)$, and we apply this to gett
$$=\int_{0}^{a} \bigg(x\frac{d}{dx}f(x)^b+f(x)^b \bigg)\mathrm dx$$
Which leads us into Winther's answer
