Problem source: problem 18, "Introduction to Real Analysis", William F. Trench, Trinity University. Here is the book
Prove that $$\int_0^1 {y^n}(1-y)^r dy = {\frac{n!}{(r + 1)(r + 2) \dots (r + n + 1)}}$$ if $n$ is a nonnegative integer and $r > -1$
I know that $$\int_0^1 y^a(1-y)^b dy = B(a + 1, b + 1) = \frac{a! b!}{(a + b + 1)!}$$ with $a! = \Gamma(a + 1)$ (or the factorial equivalent $a! = \displaystyle{\prod_{k=1}^a k}$. However, using the Beta function right away could be a little too overboard, so instead I want to solve the problem the "possibly intended" way. Define $f(n) = \displaystyle{\int_0^1 y^n(1-y)^r dy}$, we need to prove that $f(n) = \frac{n!}{\displaystyle{\prod_{k=1}^{n + 1} (r + k)}}$
Since $n$ is a nonnegative integer, we begin with the base case $n = 0$:
$$ f(0) = \int_0^1 (1-y)^r dy \stackrel{t = 1 - y, dt = -dy}{======} \int_1^0 t^r(-dt) \\ = \int_0^1 t^r dt = \left. \frac{t^{r + 1}}{r + 1} \right|_0^1 = \frac{1}{r + 1} - \frac{0}{r + 1} = \frac{1}{r + 1} = \frac{0!}{\displaystyle{\prod_{k=1}^{0+1} k}} $$
which means the statement is true with $n = 0$. Now, for some integer $n > 0$, assume the statement holds, which means $f(n) = \displaystyle{\int_0^1 y^n(1-y)^r dy} = \frac{n!}{(r + 1)(r + 2) \dots (r + n + 1)}$. We then have:
$$ f(n + 1) = \displaystyle{\int_0^1 y^{n + 1} (1-y)^r dy} = \displaystyle{\int_0^1 y \cdot y^{n} (1-y)^r dy} $$
or at least that was what I wanted, before I get stuck in this. I tried integration by parts: $$ u = y, u' = y' \\ v' = y^n(1-y)^r, v = ? $$
I can't seem to think of anything else, can you help me?
P/S: can anyone help me on how to make an extremely long equal sign with symbols on top in MathJax Latex (the type used on this site)?
You mostly have the right idea, but it's better to define $$f(n,r) = \int_0^1 y^n(1-y)^rdy$$ since the inductive step will actually change $r$.
Now we can compute $f(n,r)$ using integration by parts with $u = y^n$ and $v = -\frac{1}{r+1}(1-y)^{r+1}$. Note that $uv$ vanishes at $0$ and $1$ whenever $n>0$ and $r>-1$, so there are no boundary terms. Thus $$f(n,r) = \int_0^1 uv' = -\int_0^1 u'v = -\int_0^1 (ny^{n-1})\left(-\frac{1}{r+1}(1-y)^{r+1}\right) = \frac{n}{r+1}f(n-1,r+1).$$ From here it should be easy to inductively prove the stated formula.