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!
 A: $$\frac{1}{(1-x)^{a+1}}=\left(1+x+x^2+x^3+\ldots\right)^{a+1} $$
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.
A: Preliminary remarks: 
1) I have tried, in the following lines, to stick to the kind of answer desired by @Lui, i.e., an intuitive understanding. Of course, a rigorous demonstration is needed, but, for that, one can find explanations a little everywhere, in particular in the extraordinary book 
"Concrete Mathematics" : www.csie.ntu.edu.tw/~r97002/temp/Concrete%20Mathematics%202e.pdf  by Graham, Knuth and Patashnik (see a presentation connected to mine on page 164).
2) I have worked on $(1+x)^n$ instead of $(1-x)^n$.
Consider the usual Pascal's triangle, written in the following row-column way, with $0$ in places usually occupied by void spaces:
$$\begin{matrix}1&0&0&0&0& \ (1+x)^0\\1&1&0&0&0& \ (1+x)^1\\1&2&1&0&0& \ (1+x)^2\\1&3&3&1&0& \ (1+x)^3\\1&4&6&4&1& \ (1+x)^4\\\cdots&&&&&\end{matrix}$$
(I have placed the corresponding binomial expression on the right)
The well known building rule of Pascal's triangle says that if you take two consecutive elements in this array, their sum is to placed below the second one. Note that this rule works also for the zero coefficients.
Now, make this rule work backwards. The only possibility is to introduce negative numbers in the following way on a line above the first one (being assumed that the first element is always $1$):
$$\begin{matrix}1&-1&1&-1&1&\cdots& \ \text{coeff. of the devt of} \ (1+x)^{-1}\\1&0&0&0&0&& \ (1+x)^0\\1&1&0&0&0&& \ (1+x)^1\\1&2&1&0&0&& \ (1+x)^2\\ \cdots&&&&&&\end{matrix}$$
It works! We indeed obtain the coefficients of $\dfrac{1}{1+x}=1-x+x^2-x^3...$.
Note that we get an infinite number of non-zero coefficients... in conformity with the generalized Newton's binomial formula.
And this "backward rule" continues to work again and again...
$$\begin{matrix}\cdots&&&&&&\\1&-3&6&-10&15&\cdots& \ (1+x)^{-3}\\1&-2&3&-4&5&\cdots& \ (1+x)^{-2}\\1&-1&1&-1&1&\cdots& \ (1+x)^{-1}\\1&0&0&0&0&& \ (1+x)^0\\1&1&0&0&0&& \ (1+x)^1\\1&2&1&0&0&& \ (1+x)^2\\1&3&3&1&0&& \ (1+x)^3\\1&4&6&4&1&& \ (1+x)^4\\\cdots&&&&&&\end{matrix}$$
A: From the sum of geometric series, we know that $1/(1-x)=1+x^2+x^3+\cdots$, for $|x|<1$.
Thus, one can perceive $(1-x)^{-a-1}=1/(1-x)^{a+1}$ as the $a+1$th power of $1/(1-x)$, given $a$ is an integer,
$$
\underbrace{\frac{1}{1-x}\cdot \frac{1}{1-x} \cdots \frac{1}{1-x}}_\text{a+1} \\
=\underbrace{(1+x^2+x^3+\cdots)\times(1+x^2+x^3+\cdots)\times \cdots \times (1+x^2+x^3+\cdots)}_\text{a+1} \tag{*}
$$
From $(*)$, we know that the coefficient of $x^j$ is the number of solutions for the integer equation $m_1+m_2+\cdots+m_{a+1}=j$, with $m_k=0,1,\ldots$, which is ${a+j \choose j}$.
Therefore, we can deduce that $(1-x)^{-a-1}=\sum_{j=0}^{\infty}{{a+j} \choose j}x^j$.
