<s>[INST] Proof of binomial formula to extract coefficients of a generating function A generating function ends up rearranged into a form:
\begin{align}
(1+x+x^2+\dotsb)^n&=\frac{1}{(1-x)^n}\\[6px]
&= 1+\binom{1+n-1}{1}x+\binom{2+n-1}{2}x^2+\binom{3+n-1}{3}x^3\\
&\phantom{=\;1}+\dots+\binom{r+n-1}{r}x^r+\dotsb
\end{align}
used to extract coefficients.
I can't find a proof (some construction akin to Pascal's triangle, for example) of these coefficients working for all cases.
Probably, it is just a matter of knowing the term to include in the online search, and if this is the case, I'll be happy to delete the question.
 [/INST] The statement is obviously true for $n=1$; suppose it is for $n$; then we can see that
\begin{align}
(1+x+\dotsb)^{n+1}=\frac{1}{n}D\frac{1}{(1-x)^n}=
\frac{1}{n}\sum_{k\ge1}k\binom{k+n-1}{k}x^{k-1}=
\sum_{k\ge0}\frac{k+1}{n}\binom{k+n}{k+1}x^k
\end{align}
and it's just a matter of proving that
$$
\frac{k+1}{n}\binom{k+n}{k+1}=\binom{k+n}{k}
$$
Indeed,
$$
\frac{k+1}{n}\binom{k+n}{k+1}=
\frac{k+1}{n}\frac{(k+n)(k+n-1)\dotsm(k+n-(k+1)+1)}{(k+1)!}=
\binom{k+n}{k}
$$
</s>