Proving $(1+x)f'(x)=a f(x)$ where $f(x)=\sum_{n=0}^\infty {a\choose n}x^n$ A function $f$ is defined for $-1<x<1$ by $f(x)=\sum_{n=0}^\infty {a\choose n}x^n$. Here $a$ is a real number which is neither zero nor a positive integer. Prove that $(1+x)f'(x)=a f(x)$.
I took the derivative of $f(x)$ which is $\sum _{n=0}^\infty {a\choose n} nx^{n-1}$. And then multiplied $(x+1)$ to that and didn't find it is $a f(x)$ then I gave up and looked at the book's answer.
And in book it says:
$$f'(x)=\sum _{n=0}^\infty {a\choose n+1} (n+1)x^{n}$$. 
Now 
" $\frac {n a(a-1)...(a-n+1)}{n!}+\frac{(n+1)a(a-1)...(a-n)}{(n+1)!}=\frac{a(a-1)...(a-n+1)} {(n-1)!}(1+\frac{(a-n)}{n})$.
So $(1+x)f'(x)=a f(x)$." This is the part where I don't understand. I don't know how after 
writing this $\frac {n a(a-1)...(a-n+1)}{n!}+\frac{(n+1)a(a-1)...(a-n)}{(n+1)!}=\frac{a(a-1)...(a-n+1)} {(n-1)!}(1+\frac{(a-n)}{n})$ you can know that $(1+x)f'(x)=a f(x)$ is true. Can anyone explain?

For problems like these, I like to write out the first few terms in order to really understand what I've got.
We are working with $f$ defined by
$$f(x) = \binom a0 + \binom a1 x + \binom a2 x^2 + \cdots.$$
Your definition of $f'(x)$ is correct, except that the series should start at $n=1$ since the derivative of $\binom a0$ is simply $0$. Then,
$$f'(x) = 1\binom a1 + 2\binom a2 x + 3\binom a3 x^2 + \cdots = \sum_{n=1}^\infty n\binom{a}{n} x^{n-1}.$$
Now, in general, we have that
$$n \binom an = n \cdot \frac{a!}{n! (a-n)!} = n \cdot \frac{a \cdot (a-1)!}{n \cdot (n-1)! (a-n)!} = a \cdot \frac{(a-1)!}{(n-1)! (a-1 - (n-1))!} = a \binom{a-1}{n-1}.$$
Substituting this into our definition of $f'(x)$ shows that
$$f'(x) = a\binom {a-1}0 + a\binom {a-1}1 x + a\binom {a-1}2 x^2 + \cdots = \sum_{n=1}^\infty a\binom{a-1}{n-1} x^{n-1} = \sum_{n=0}^\infty a\binom{a-1}{n} x^{n}.$$
The last step above stems from the transformation $n \mapsto n+1$.
Now, there is another identity about binomials which is well-known, namely Pascal's Identity, which states that
$$\binom{a-1}n + \binom{a-1}{n-1} = \binom an.$$
Note that
$$\begin{align*}
f'(x) + xf'(x) &= \left(a\binom {a-1}0 + a\binom {a-1}1 x + a\binom {a-1}2 x^2 + \cdots\right) \\& \quad+ \left(a\binom {a-1}0x + a\binom {a-1}1 x^2 + \cdots\right), \\
&= a\binom {a-1}0 + \left(a\binom {a-1}0 + a\binom{a-1}1\right)x + \left(a\binom {a-1}1 + a\binom{a-1}2\right) x^2 + \cdots, \\
&= a\left(\binom {a-1}0 + \binom{a}1x + \binom{a}2 x^2 + \cdots\right), \\
&= af(x).
\end{align*}$$
In the above, we have $\binom{a-1}0 = \binom a0 = 1$, so our reasoning is valid. Hence, $f'(x) + xf'(x) = af(x)$, or
$$(1+x)f'(x) = af(x).$$
