Integral of polynomial I want so compute the following integral for $S \in \mathbb{R}_+$ and $k,n \in \mathbb{N}$
$$
V(S,k,n) := \int_0^S (S-x)^k x^n \, d x.
$$
According to Mathematica the solution is
\begin{equation}
V(S,k,n) = S^{k+n+1} \frac{k! n!}{(k+n+1)!} .
\end{equation}
How can that be proven? Expanding $(S-x)^k = \sum_{j=0}^k \binom{k}{j}(-x)^k S^{k-j}$ is my first step, but then I get stuck and I guess one needs a tricky identity regarding alternating binomial coefficients?
 A: Given $\displaystyle V(S,k,n)= \int^{S}_{0}(S-x)^kx^ndx,$ Put $x=St\;,$ Then $dx = Sdt$ and changing imits
So $\displaystyle V(S,k,n) = S^{k+n+1}\int^{1}_{0}(1-t)^{k}t^{n}dt = S^{k+n+1}I_{k,n}$
So $\displaystyle I_{k,n} = \int^{1}_{0}(1-t)^{k}t^{n}dt = \int^{1}_{0}t^{k}(1-t)^{n}dt=\int^{1}_{0}x^{k}(1-x)^{n}dx$
$\displaystyle I_{k,n}=\int_{0}^{1}x^k(1-x)^n$ d$x $ Using IBP, $$\displaystyle \\ I_{k,n}= \frac{(1-x)^nx^{k+1}}{k+1}|_{0}^{1}+\int_{0}^{1}\frac{x^{k+1}}{k+1}\cdot n(1-x)^{n-1}dx$$
Using IBP again, and again, etc, it is clear that $$ \\\\I_{k,n}= \frac{(1-x)^nx^{k+1}}{k+1}|_{0}^{1}+\frac{n(1-x)^{n-1}x^{k+2}}{(k+1)(k+2)}|_{0}^{1}+\frac{n(n-1)(1-x)^{n-2}x^{k+3}}{(k+1)(k+2)(k+3)}|_{0}^{1}+...+\frac{n!x^{k+n+1}}{(k+1)(k+2)(k+3)..(k+n+1)}|_{0}^{1} + \frac{n!0(1-x)^{-1}x^{k+n+2}}{(k+1)(k+2)..(k+n+2)}|_{0}^{1}+.. \\ \\ $$ 
Evaluating the boundaries, all the terms except one vanishes, thus $$\displaystyle \\ I_{k,n}=\frac{n!}{(k+1)(k+2)(k+3)..(k+n+1)} =\frac{k!n!}{(k+n+1)!}$$
So $$ I_{k,n} = \int^{1}_{0}(1-t)^{k}t^{n}dt = \int^{1}_{0}t^{k}(1-t)^{n}dt = S^{k+n+1}\cdot \frac{k!\cdot n!}{(k+n+1)!}$$
A: Let $y=x/S$
\begin{align}
\int\limits_{0}^{S} (S-x)^{k} x^{n} \mathrm{d}x &= \int\limits_{0}^{1} (S-Sy)^{k} S^{n} y^{n} S \mathrm{d}y \\
&= S^{k+n+1}\int\limits_{0}^{1} (1-y)^{k} y^{n} \mathrm{d}y \\
&= S^{k+n+1} \mathrm{B}(n+1,k+1) \\
&= S^{k+n+1} \frac{\Gamma(n+1)\Gamma(k+1)}{\Gamma(n+k+2)} \\
&= S^{k+n+1} \frac{n! k!}{(n+k+1)!}
\end{align}
A: Binomial expansion was a good idea but there is a mistake in your formula is $(-x)^j$ :
$$V(S,k,n)=\int_{0}^{S}(S-x)^kx^ndx$$
$$=\int_{0}^{S}\sum_{j=0}^{k}\binom{k}{j}(-x)^jS^{k-j} x^ndx$$
$$=\sum_{j=0}^{k}\binom{k}{j}S^{k-j}(-1)^j\int_{0}^{S} x^{n+j}dx$$
$$=\sum_{j=0}^{k}\binom{k}{j}S^{k-j}(-1)^j\frac{S^{n+j+1}}{n+j+1}$$
$$=\sum_{j=0}^{k}\binom{k}{j}(-1)^j\frac{S^{n+k+1}}{n+j+1}$$
$$=S^{n+k+1}\sum_{j=0}^{k}\binom{k}{j}(-1)^j\frac{1}{n+j+1}$$
$$=S^{n+k+1}k!\sum_{j=0}^{k}\frac{1}{j!(k-j)!}(-1)^j\frac{1}{n+j+1}$$
