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.