Intuitive explanation of $(1-x)^{-a-1}=\sum_{j=0}^{\infty}{{a+j} \choose j}x^j$ Could anyone please explain me the reasoning behind this formula?
$(1-x)^{-a-1}=\sum_{j=0}^{\infty}{{a+j} \choose j}x^j$
Thanks so much!

$$\frac{1}{(1-x)^{a+1}}=\left(1+x+x^2+x^3+\ldots\right)^{a+1} $$
hence the coefficient of $x^n$ in this power series is the number of elements of the set:
$$ E_n = \{(n_1,n_2,\ldots,n_{a+1}):\, n_i\in\mathbb{N},\, n_1+n_2+\ldots+n_{a+1}=n \} $$
that by stars and bars ig given by $\binom{n+a}{a}$. It follows that:
$$ \frac{1}{(1-x)^{a+1}}=\sum_{n\geq 0}\binom{n+a}{a}x^n $$
as wanted. As an alternative, we may just differentiate
$$ \frac{1}{1-x} = \sum_{n\geq 0} x^n $$
multiple times.
