Prove \(\int_{0}^{1} (1-x^{1/k})^{n} dx = {1\over {k+n\choose n}}\)

How to prove
\begin{equation}
\int\limits_{0}^{1} (1-x^{1/k})^{n} dx = {1\over {k+n\choose n}}
\end{equation}
was a question posed at Mathematics Stack Exchange. Here is my solution.

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}

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