This is essentially an exercise in integration. If $X$ has a Beta($\alpha$,$\beta$) distribution then the expected value you're trying to calculate can be written as:
$ \int_{0}^{1} x^r (1-x)^s \frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )}x^{\alpha -1}(1-x)^{\beta -1} dx = \int_{0}^{1}\frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )}x^{r+\alpha -1}(1-x)^{s+\beta -1} dx $
The trick is to multiply by constants so that the integrand is another beta density, which we know integrates to 1. The constants required will all be $\Gamma$ functions:
$ E(X^r (1-X)^s ) = \frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )} \cdot \frac{\Gamma (\alpha+r )\Gamma (\beta+s )}{\Gamma (\alpha +\beta+r+s )}\int_{0}^{1} \underbrace{\frac{\Gamma (\alpha +\beta+r+s )}{\Gamma (\alpha+r )\Gamma (\beta+s )} x^{r+\alpha -1}(1-x)^{s+\beta -1}}_{{\rm Beta}(\alpha+r, \beta+s) \ \ {\rm density}}dx $
So the integral equals 1, since it's a probability density. Therefore,
$ E(X^r (1-X)^s ) = \frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )} \cdot \frac{\Gamma (\alpha+r )\Gamma (\beta+s )}{\Gamma (\alpha +\beta+r+s )} $
I hope this helps!