proving for $\int_{0}^{1}x^{k}\cdot \left(1-x\right)^{n-k}dx = \frac{1}{\binom{n}{k}\cdot (n+1)}$ How can we prove $\displaystyle \int_{0}^{1}x^{k}\cdot \left(1-x\right)^{n-k}dx = \frac{1}{\binom{n}{k}\cdot (n+1)}$
$\bf{My\; Try}::$ Let $\displaystyle I_{k,n-k} = \int_{0}^{1}\left(1-x\right)^{n-k}\cdot x^kdx = \left[(1-x)^{n-k}\cdot \frac{x^{k+1}}{k+1}\right]_{0}^{1}-\frac{\left(n-k\right)}{k+1}\int_{0}^{1}(1-x)^{n-k-1}\cdot x^{k+1}dx$
$\displaystyle I_{k,n-k}= -\frac{\left(n-k\right)}{\left(k+1\right)}\cdot I_{k+1,n-k-1}$
Using  Recursively we get
$\displaystyle I_{k,n-k} = -\frac{\left(n-k\right)}{\left(k+1\right)}\cdot I_{k+2,n-k-2} = +\frac{(n-k)(n-k-1)}{(k+1)(k+2)}\cdot I_{k+3,n-k-3} = $
Now I did  not understand how can i solve after that
help required.
Thanks
 A: The basic idea of identifying the recursion via integration by parts is fine. You seem to have mixed up your indices a bit, however.
Nevertheless, in particular since you know where you want to arrive, I would suggest doing it in the form
$$\begin{align}
\frac{n!}{k!(n-k)!} \int_0^1 x^k(1-x)^{n-k}\,dx &= \frac{n!}{(k+1)!(n-k)!} \left[x^{k+1}(1-x)^{n-k}\right]_0^1\\
&\quad + \frac{n!}{(k+1)!(n-k-1)!}\int_0^1 x^{k+1}(1-x)^{n-k-1}\,dx\\
&=  \frac{n!}{(k+1)!(n-k-1)!}\int_0^1 x^{k+1}(1-x)^{n-k-1}\,dx\\
&= \quad \dotsb\\
&= \frac{n!}{n!(n-n)!} \int_0^1 x^n(1-x)^{n-n}\,dx\\
&= \int_0^1 x^n\,dx\\
&= \frac{1}{n+1}.
\end{align}$$
A: This is the Beta Function: assuming $\;n,k\in\Bbb N\;$
$$I_k:=\int\limits_0^1 x^k(1-x)^{n-k}dx=:B(k+1,n-k+1)=$$
$$=\frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+2)}=\frac{k!(n-k)!}{(n+1)!}$$
Or directly integrating by parts:
$$u=x^k\;\;,\;\;u'=kx^{k-1}\\v'=(1-x)^{n-k}\;\;,\;\;v=-\frac1{n-k+1}(1-x)^{n-k+1}\;\;\implies$$
$$I_k=\left.-\frac1{n-k+1}x^k(1-x)^{n-k+1}\right|_0^1+\frac k{n-k+1}\int\limits_0^1 x^{k-1}(1-x)^{n-k+1}dx=$$
$$=\frac k{n-k+1}I_{k-1}$$
and now a simple inductive argument wraps this up.
