Show that $\displaystyle \displaystyle \sum_{k=0}^{n}\frac{\alpha^{k+1}}{k+1}\binom{n}{k} = \frac{(\alpha+1)^{n+1}-1}{n+1}$.
$\displaystyle \displaystyle (1+\alpha)^n = \sum_{k=0}^{n}\binom{n}{k}\alpha^k$, therefore $\displaystyle \displaystyle \int(1+\alpha)^n \;{d\alpha} = \int\sum_{k=0}^{n}\binom{n}{k}\alpha^k\;{d\alpha}.$
So $\displaystyle \displaystyle \frac{(1+\alpha)^{n+1}}{n+1} + C = \sum_{k=0}^n \frac{\alpha^{k+1}}{k+1}\binom{n}{k}. $ Put $\displaystyle \alpha = 0$, to get $\displaystyle C = -\frac{1}{n+1}$.
Thus $\displaystyle \displaystyle \sum_{k=0}^{n}\frac{\alpha^{k+1}}{k+1}\binom{n}{k} = \frac{(\alpha+1)^{n+1}}{n+1}-\frac{1}{n+1} = \frac{(\alpha+1)^{n+1}-1}{n+1}.$