Understanding the binomial expansion for negative and fractional indices? I have been trying to understand why the binomial theorem can work for negative and fractional indices.
I understand that when raising binomials to positive integral indices, each coefficient is simply the number of ways that you can pick each term (e.g. for $(x+y)^5$, if you want to make up $xy^4$ there are 5 brackets from which to pick the $x$, so this term will come up 5 times in the full expansion).
I am not sure if a way to understand the infinite expansion for negative and fractional indices exists, but if it does I would very much like to know! Otherwise, I haven't been able to find a proof that shows that the result of the expansion for positive integer powers is valid for negative or fractional indices. I would be very grateful if someone could point me in the direction of such a proof.
 A: I don't know if this is what you are looking for but I feel like it warrants a mention, if you do not understand binomial expansion for indices say $n \lt 0$ then it helps to think of it in this way. Note, this is only for $n \in \mathbb{I}$.  Anyway I am assuming you know that we define an infinite geometric progression $A_n$ like this, $$a + ar + ar^2 + ar^3 + \ldots + ar^{n} + \ldots \infty$$ Now if we define the sum of this progression $$\sum ^{\infty} _ 1 A_n = S$$ then we can say that $$\sum ^{\infty} _ 1 A_n = S = \dfrac{a}{1-r}$$ Now, let us say the first term $a$ is $1$ and the common ratio $r=x$ then we get, the series as $$A_x = 1 + x + x^2 + x^3 + \ldots + x^{n} + \ldots \infty$$ and we can define the sum as, $$\sum ^{\infty} _ 1 A_x = S = \dfrac{1}{1-x}$$ which can be written as, $$\left(1-x\right)^{-1}$$ so, you can now see what really goes on using an $\infty$ geometric progression $A_x$.
For higher powers obviously, you can define it as $$\left(1 + x + x^2 + x^3 + \ldots + x^{n} + \ldots \infty \right) \cdot \left(1 + x + x^2 + x^3 + \ldots + x^{n} + \ldots \infty \right)= \left(1-x\right)^{-2}$$ and for a concise formula you can refer to the Taylor series, which has been discussed in the previous question.  Obviously if we wanted to do this for any expansion we could define it using a geometric progression by changing the common ratio. For the fractional indices, $n \in \mathbb{R}$ you can refer to the proof given above. 
A: This is probably the wrong proof for you, but I will post it anyways.  (requires calculus)
Note that $f(x)=(a+x)^n$ is an analytic function in $x$ for arbitrary $a,n$ since on its own, it is a power series with one term.
If it is an analytic function, then it should follow Taylor's theorem.
Now, if we take the expansion around $x=0$, we get
$$(a+x)^n=a^n+na^{n-1}x+\frac{n(n+1)}2a^{n-2}x^2+\dots$$
Since $f(0)=a^n$, $f'(0)=na^{n-1}$, $\dots f^{(k)}(0)=n(n+1)(n+2)\dots(n+k-1)a^{n-k}$
or
$$(a+x)^n=\sum_{k=0}^\infty\frac{n(n+1)(n+2)\dots(n+k-1)}{k!}a^{n-k}x^n$$
$$(a+x)^n=\sum_{k=0}^\infty\binom nka^{n-k}x^n$$
where $f'(x)$ is the first derivative of $f(x)$, $f''(x)$ the second derivative, etc. $f^{(k)}(x)$ is the $k$th derivative of $f(x)$.
