Show that:$\int_{0}^{1}(1-\sqrt[k]{x})^ndx={1\over {k+n\choose n}}$ Show that 
$$I=\int_{0}^{1}(1-\sqrt[k]{x})^ndx={1\over {k+n\choose n}}$$
My try:
$x=u^k$ then $dx=ku^{k-1}du$
$$I=k\int_{0}^{1}(1-u)^n u^{k-1}du$$
$v=1-u$ then $dv=-du$
$$I=k\int_{0}^{1}v^n(1-v)^{k-1}dv$$
Using the binomial theorem to expand and integrate term by term is tedious, can anyone show me another quick easy way please? Thank you.
 A: Let $y=x^{1/k}$
\begin{align}
\int\limits_{0}^{1} (1-x^{1/k})^{n} dx &= k\int\limits_{0}^{1} (1-y)^{n} y^{k-1} dy \\
&= k \mathrm{B}(k,n+1) \\
&= \frac{k\Gamma(k)\Gamma(n+1)}{\Gamma(k+n+1)} \\
&= \frac{k!n!}{(k+n)!} \\
&= {1\over {k+n\choose n}}
\end{align}
We have used the beta and gamma functions.
A: By integration by parts,
$$ B(a,b) = \int_{0}^{1}x^{a-1}(1-x)^{b-1}\,dx = \frac{\Gamma(a)\,\Gamma(b)}{\Gamma(a+b)}\tag{1} $$
so if $m$ and $n$ are natural numbers
$$ \int_{0}^{1}x^m(1-x)^{n-1}\,dx = \frac{\Gamma(m+1)\Gamma(n)}{\Gamma(m+n+1)} = \frac{m!(n-1)!}{(m+n)!} = \frac{1}{n\binom{m+n}{n}}.\tag{2}$$
Have a look at Euler's Beta function.
A: Maybe you can try the Laplace transform. Let $f(t)=\int_0^t \tau^a(t-\tau)^b d\tau$, it's easy to compute $\mathcal{L}(f)(s)=\frac{a!b!}{s^{a+b+2}}$, then $f(t)=\frac{a!b!}{(a+b+1)!} t^{a+b+1}$.
A: First of all this is true for $n=1$.
Let's assume that it is true for $n=s-1$. Now we proceed to show that this is true for $n=s$.
Now $$I_s=\int_{0}^{1}(1-\sqrt[k]{x})^sdx=x(1-\sqrt[k]{x})^s|_0^1+\frac{s}{k}\int_0^1 x (1-\sqrt[k]{x})^{s-1} x^{\frac{1-k}{k} } dx=\frac{s}{k}\int_{0}^{1}x^{\frac{1}{k}}(1-\sqrt[k]x)^{s-1}=-\frac{s}{k}\int_{0}^{1}-x^{\frac{1}{k}}(1-\sqrt[k]x)^{s-1}=-\frac{s}{k}\int_{0}^{1}(1-x^{\frac{1}{k}})(1-\sqrt[k]x)^{s-1}+\frac{s}{k}I_{s-1}=-\frac{s}{k}I_s+\frac{s}{k}I_{s-1}$$
Then $$I_s \frac{s+k}{k}=\frac{s}{k}{1\over {k+s-1\choose s-1}}$$
Thus $$I_s=\frac{s}{s+k}\times \frac{(s-1)!k!}{(k+s-1)!}=\frac{s!k!}{(s+k)!}=\frac{1}{s+k \choose s}$$
Thus, This is true for all $n$
