Integration Problem how can I prove this identity:
$$
n \in N_0, m \in N
$$
$$
\int_0^1 x^{n}(1-x)^{m}dx\ = \frac{n!m!}{(n+m+1)!}
$$
I thought about the geometric series but I am not sure
Thanks in advance!
 A: Let $I_{n,m}$ be the integral in question.
Integration by parts with $u=x^n$, $dv=(1-x)^m\,dx$, $du=nx^{n-1}\,dx$, $v=-\frac{(1-x)^{m+1}}{m+1}$ yields
$$
\begin{align*}
I_{n,m}&=\int_0^1x^n(1-x)^m\,dx\\
&=\left.-\frac{x^n(1-x)^{m+1}}{m+1}\right\vert_{x=0}^{x=1}+\frac{n}{m+1}\int_0^1x^{n-1}(1-x)^{m+1}\,dx\\
&=\frac{n}{m+1}I_{n-1,m+1}.
\end{align*}
$$
Applying this iteratively, we see that
$$
\begin{align*}
I_{n,m}&=\frac{n}{m+1}\cdot I_{n-1,m+1}\\
&=\frac{n}{m+1}\cdot\frac{n-1}{m+2}\cdot I_{n-2,m+2}\\
&=\ldots\\
&=\frac{n}{m+1}\cdot\frac{n-1}{m+2}\cdots\frac{1}{m+n}\cdot I_{0,m+n}\\
&=\frac{n!\cdot m!}{(m+n)!}\cdot I_{0,m+n}
\end{align*}
$$
Now, can we compute $I_{0,m+n}$?
$$
\begin{align*}
I_{0,m+n}&=\int_0^1(1-x)^{m+n}\,dx\\
&=\left.-\frac{(1-x)^{m+n+1}}{m+n+1}\right\rvert_{x=0}^{x=1}\\
&=\frac{1}{m+n+1}.
\end{align*}
$$
So, in all, we get
$$
I_{n,m}=\frac{n!\cdot m!}{(n+m)!}\cdot\frac{1}{m+n+1}=\frac{n!\cdot m!}{(n+m+1)!},
$$
as desired.
A: Note that the beta function can be defines as $$B(x,y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}dy = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} = \frac{(x-1)!(y-1)!}{(x+y -1)!}$$ Note that $$\Gamma(n) = (n-1)!$$ and $$\Gamma(n+1) = n\Gamma(n)$$ In your case we have $$\int_{0}^{1}x^n (1-x)^{m}dx$$
Thus applying the formula above we have 
$$\int_{0}^{1}x^n (1-x)^{m}dx = \frac{\Gamma(n+1)\Gamma(m+1)}{\Gamma(n+m+1)}$$
You should be able to continue from here.
