= Power-complement integral recurrence
{title2=$J_n(p)=\int_0^1(1-s^p)^n\,ds$}
For $p>0$ and a nonnegative integer $n$, <integration by parts> gives, when $n>0$,
$$
J_n(p)=pn\int_0^1s^p(1-s^p)^{n-1}\,ds
=pn\bigl(J_{n-1}(p)-J_n(p)\bigr).
$$
The endpoint term $[s(1-s^p)^n]_0^1$ vanishes. Therefore
$$
J_n(p)=\frac{pn}{pn+1}J_{n-1}(p),\qquad J_0(p)=1.
$$
Repeated application yields $J_n(p)=\prod_{j=1}^n pj/(pj+1)$. The <integration by substitution> $w=s^p$ also identifies $J_n(p)=B(1/p,n+1)/p$ with an <Euler beta function>. A related weighted <integral> follows immediately from the same substitution:
$$
\int_0^1s^{p-1}(1-s^p)^n\,ds=\frac1{p(n+1)}.
$$
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