Binomial expansion with gamma function

Question

I'm studying Euler proof of $\zeta(2)$. There are several proof but I'm on one. I found in a book this expression for binomial but I can't find where it comes from

$(1+y)^{-t} = \frac{1}{\Gamma(t)} \sum_{n=0}^{\infty} (-1)^{n} \frac{\Gamma(t+n)}{n!} y^n$.

Does some know where it comes from.

Answer 1

This binomial theorem is the Maclaurin series of $(1+y)^{-t}$. You may prove by induction that the $n$th derivative of this function is $(-1)^n\frac{\Gamma(t+n)}{\Gamma(t)}(1+y)^{-t-n}$, which gives the desired coefficients when we expand around $y=0$. The remainder term's modulus has an upper bound proportional to $y^{n+1}$ (see here for details), so the series is valid if $|y|<1$.