Express $$\int_0^1x^m(1-x^n)^pdx$$ in terms of gama function and hence evaluate the integral.
I used the substitution $x^n=y$ and solving got this integral as equal to the beta-function $${1\over n}\beta({{{m+1}\over n},{p+1}})$$ solving futher by converting to gamma function ($\gamma)$ i get
$${1 \over n}[{{\Gamma({m+1 \over n})\Gamma(p+1)}} \over \Gamma({{m+1} \over n}+p+1)$$
and the answer i finally get is
$$p\over (m+p-1)!(m+np+1)$$
Is it correct ? One doubt i had in the solution is can we write
$$\Gamma(a+p)={{(a+p-1)!} \over (a-1)!} \Gamma(a)$$ ??
$$ \begin{align} \int_0^1x^m(1-x^n)^p\,\mathrm{d}x &=\int_0^1u^{m/n}(1-u)^p\,\mathrm{d}u^{1/n}\\ &=\frac1n\int_0^1u^{(m+1)/n-1}(1-u)^p\,\mathrm{d}u\\ &=\frac1n\frac{\Gamma\left(\frac{m+1}{n}\right)\Gamma(p+1)}{\Gamma\left(\frac{m+1}{n}+p+1\right)}\\ &=\frac{p}{m+1+pn}\frac{\Gamma\left(\frac{m+1}{n}\right)\Gamma(p)}{\Gamma\left(\frac{m+1}{n}+p\right)} \end{align} $$ So your initial answer looks good. However, I don't see how you get from there to $$ \frac{p}{(m+p-1)!(m+np+1)} $$
In the last question, for integer $n$, $\Gamma(n)=(n-1)!$. The rest is simply cancellation.