Proving an integral identity involving binomial coefficients I'm having a lot of trouble proving or disproving the following statement:
$$\int_0^1(1-x^{\alpha})^n\ dx=\frac{n!\alpha^n}{\prod_{k=1}^n(\alpha k+1)}$$
$\alpha\in\mathbb{R}^{+*}$ and $n\in\mathbb{N}^*$. I've tried proving it via the binomial coefficients and an induction, but this has lead to nothing:
$$\int_0^1(1-x^{\alpha})^n\ dx=\int_0^1 \sum_{k=0}^n\begin{pmatrix}
n\\
k
\end{pmatrix}(-x^\alpha)^k\ dx=\sum_{k=0}^n\begin{pmatrix}
n\\
k
\end{pmatrix}(-1)^k\int_0^1x^{\alpha k}\ dx$$
$$=\sum_{k=0}^n\begin{pmatrix}
n\\
k
\end{pmatrix}(-1)^k(\alpha k+1)^{-1}$$
The induction blocks beccause of the fact that:
$$\begin{pmatrix}
n\\
k
\end{pmatrix}=\frac{n!}{k!(n-k)!}$$
Making it impossible to factor out $(n-k+1)$ from the denominator in the hereditary step of the proof because this term isn't constant.
Attacking the problem via series has also proven unfruitful on my end; the series can't be evaluated at $x=1$ or $x=0$, and evaluating it at other points leads to a disgusting series. I'm just hoping someone can give me some indicators on what to do!
This is a purely recereational question and I thought it would be interesting to tie this conjecture to integrals of this form as I believe there was a similar method for deriving the Wallis product for $\pi$ via integrals like this.
For the reference to this integral and the wallis product for $\pi$, see the "Historical motivation" paragraph of this article:
https://mindyourdecisions.com/blog/2016/10/12/the-wallis-product-formula-for-pi-and-its-proof/
 A: We seek to show that
$$\int_0^1 (1-x^\alpha)^n \; dx
= \frac{n! \alpha^n}{\prod_{k=1}^n (\alpha k + 1)}.$$
We get for the LHS
$$\sum_{k=0}^n {n\choose k} (-1)^k
\int_0^1 x^{\alpha k} \; dx
= \sum_{k=0}^n {n\choose k} (-1)^k
\frac{1}{\alpha k + 1}.$$
Now introduce
$$f(z) = n! (-1)^n \frac{1}{z\alpha+1}
\prod_{q=0}^n \frac{1}{z-q}.$$
We have with $0\le k\le n$ that
$$\mathrm{Res}(f(z); z=k)
= n! (-1)^n \frac{1}{\alpha k + 1}
\prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k-q}
\\ = n! (-1)^n \frac{1}{\alpha k + 1}
\frac{1}{k!} (-1)^{n-k} \frac{1}{(n-k)!}
= {n\choose k} (-1)^k \frac{1}{\alpha k + 1}.$$
These are precisely the terms of our sum. Now residues sum to zero and
the residue at infinity is zero, leaving for the sum
$$- \mathrm{Res}(f(z); z=-1/\alpha)
= - \frac{n! (-1)^n}{\alpha} \prod_{q=0}^n \frac{1}{-1/\alpha-q}
\\ = - n! (-1)^n \alpha^n \prod_{q=0}^n \frac{1}{-1-\alpha q}
= n! \alpha^n \prod_{q=0}^n \frac{1}{\alpha q + 1}.$$
This is the claim. With the poles of $f(z)$ at the integers from the
range $[0,n]$ and $\alpha>0$ there is no co-incidence with the pole at
$z = -1/\alpha.$
A: Using the Beta function, we can proceed as follows. Make the substitution $u=  x^{\alpha}$ so that
$$
I =\int_0^1(1-x^\alpha)^n\,dx =\frac{1}{\alpha}\int_0^1(1-u)^n u^{1/\alpha-1}\,du =
\frac{\Gamma(n+1)\Gamma(\alpha^{-1})}{\alpha\Gamma(n+1+\alpha^{-1})}\tag{0}
$$
Using the identity $\Gamma(a+1) = a\Gamma(a)$ for $a>0$ and $\Gamma(1) = 1$ we find that $\Gamma(n+1) = n!$ and
\begin{align}
\Gamma(n+1+\alpha^{-1}) & =(\alpha^{-1}+n) (\alpha^{-1}+n-1)\dotsb(\alpha^{-1}+1)(\alpha^{-1})\Gamma(\alpha^{-1})\\
 & = \frac{\prod_{k=1}^n (\alpha k+1) \Gamma(\alpha^{-1})}{\alpha^{n+1}} .
\end{align}
Substituting into $(0)$ we get that
$$
I=\frac{n!\alpha^n}{\prod_{k=1}^n (\alpha k+1) }
$$
