This question is about equation number (4) in Philip J. Davis’ Essay titled "LEONHARD EULER'S INTEGRAL: A HISTORICAL PROFILE OF THE GAMMA FUNCTION".
In there it is stated by the author "Euler expanded $(1-x)^n$ by the binomial theorem, and without difficulty found that:"
$$ \int_0^1 x^e (1-x)^n dx = \frac{1·2···n}{(e+1)(e+2)···(e+n+1)}$$
In this case $e$ is an arbitrary value and $n$ is an integer. I tried to reproduce this by myself but I get stuck with the coefficients of the binomial, I do not see how to cancel them nicely to arrive at this solution by Euler.
Could somebody help out understanding Euler's work by providing a sketch on how to arrive to this equality?
I have not read Euler's original work, but I assume it went like this:
We have
\begin{align} \int_0^1 x^e (1-x)^n dx &= \int_0^1 x^e \sum_{k=0}^n \binom{n}{k} (-x)^kdx = \\ &= \sum_{k=0}^n \binom{n}{k} (-1)^k \int_0^1 x^{e+k} dx = \\ &= \sum_{k=0}^n \frac{n!}{k! (n-k)!} \frac{(-1)^k}{e+k+1} = \\ &= \frac{p(e)}{(e+1)(e+2)\dots(e+n+1)}\end{align} where $p(e)$ is a certain polynomial of degree $n$, specificaly $$ p(e) = \sum_{k=0}^n \frac{n!}{k! (n-k)!} (-1)^k \prod_{i\in\{0,\dots,n\}\setminus\{k\}} (e+i+1) $$ If we put $e=-j-1$, $j\in\{0,\dots, n\}$ only one term in the sum remains and we get \begin{align} p(-j-1) &= \frac{n!}{j! (n-j)!} (-1)^j \prod_{i\in\{0,\dots,n\}\setminus\{j\}} (i-j) = \\ &= \frac{n!}{j! (n-j)!} (-1)^j \cdot (-1)^j j!(n-j)! = \\ &= n!\end{align} Since $p(e)$ takes value $n!$ in $n+1$ different points, and $p(e)$ is a polynomial of degree $n$, that means that $p(e) =n!$. Thus we get $$ \int_0^1 x^e (1-x)^n dx = \frac{n!}{(e+1)(e+2)\dots(e+n+1)}$$