Prove by induction that$ \int_{0}^{1} y^n(1-y)^r dy = \frac {n!}{(r+1)(r+2)...(r+n+1)} $ 
Prove by induction that
$$ \int_{0}^{1} y^n(1-y)^r dy = \frac {n!}{(r+1)(r+2)...(r+n+1)} $$ if $n$ is a non-negative and $r>-1$

Testing n = 1:
$$ \int_{0}^{1} y(1-y)^r dy = \frac {1}{(r+1)(r+2)} $$
I am new to this. I am reviewing my courses on Calculus (years ago), But I am literally stock.  What would the approach be here?
Much appreciated
 A: Off the top of my head,
this might help:
$\begin{array}\\
I(n, r+1)
&=\int_{0}^{1} y^n(1-y)^{r+1} dy \\
&=\int_{0}^{1} y^n(1-y)(1-y)^r dy\\
&=\int_{0}^{1} y^n(1-y)^r dy-\int_{0}^{1} y^{n+1}(1-y)^r dy\\
&=I(n, r)-I(n+1, r)\\
\end{array}
$ 
with
$I(n, 0)
=\int_{0}^{1} y^n dy
=\dfrac1{n+1}
$.
A: Integration by parts:
For $n=1$,
$$\int_0^1 y(1-y)^r \mathop{dy} = \left[-\frac{y}{r+1}(1-y)^{r+1}\right]_{y=0}^1 + \frac{1}{r+1} \int_0^1 (1-y)^{r+1} \mathop{dy} = \frac{1}{(r+1)(r+2)}.$$
[Note that the last step is similar to how you would show the $n=0$ case.]

General case:
If it holds for $n-1$, then verifying the case for $n$:
\begin{align}
\int_0^1 y^n (1-y)^r \mathop{dy}
&= \left[-\frac{y^n}{r+1}(1-y)^{r+1}\right]_{y=0}^1
+ \frac{n}{r+1} \int_0^1 y^{n-1} (1-y)^{r+1} \mathop{dy}
\\
&= \frac{n}{r+1} \cdot \frac{(n-1)!}{(r+2)(r+3) \cdots (r+n)}
\end{align}
A: HINT:
Integrate by parts,
$$I_{(n,r)}=\int_0^1y^n(1-y)^rdy=\cdots\dfrac n{r+1}I_{(n-1,r+1)}$$
$$=\dfrac{n(n-1)}{(r+1)(r+2)}I_{(n-2,r+2)}=\cdots=\dfrac{n!}{\prod_{u=1}^n(r+u)}I_{(0,r+n)}$$
