Proving Newton's Binomial Theorem So, I've done most of the problem to this point, but just cannot figure out the last piece.  I may just be missing the math skills needed to complete the proof (differential equations).
Problem (from Rudin): If $\alpha$ is real and $-1 < x < 1$, prove Newton's binomial theorem; 
$(1+x)^\alpha$ = $1 + \sum{\frac{\alpha (\alpha-1) \dots(\alpha - n + 1)}{n!}}$$x^n$
Where the sum goes from $n=1$ to infinity.  The book suggests calling the right side $f(x)$, proving the series converges, proving $(1+x)f'(x) = \alpha f(x)$. I've done all this.  All that is left is solving this differential equation, which I simply cannot figure out.  Could anyone help with this last step?  It would be appreciated!
 A: To settle this: there is the formula for logarithmic differentiation,
$$\frac{\mathrm d}{\mathrm dx}\log(f(x))=\frac{f^\prime(x)}{f(x)}$$
which means
$$\frac{\mathrm d}{\mathrm dx}\log(f(x))=\frac{\alpha}{1+x}$$
Integrate both sides, mind the boundary condition, and you should be able to recover your needed result.
A: For compactness, use Knuth's notation for falling factorial powers:
$$
\alpha^{\underline{n}} 
  = \alpha \cdot (\alpha - 1) \cdot \ldots \cdot (\alpha - n + 1)
  = \prod_{0 \le k \le n - 1} (\alpha - k)
$$
Now use Maclaurin's formula:
$$
\begin{align*}
f(z) &= (1 + z)^\alpha \\
f^{(n)}(x) &= \alpha^{\underline{n}} (1 + z)^{\alpha - n}
\end{align*}
$$
This gives directly:
$$
f(z) = \sum_{n \ge 0} \frac{f^{(n)}(0)}{n!} z^n
     = \sum_{n \ge 0} \frac{\alpha^{\underline{n}}}{n!} z^n
$$
We can define:
$$
\binom{\alpha}{n} = \frac{\alpha^{\underline{n}}}{n!}
$$
to get the familiar:
$$
(1 + z)^\alpha
  = \sum_{n \ge 0} \binom{\alpha}{n} z^n
$$
A: Note that the differential equation $$f'-\frac{\alpha}{1+x}f=0$$ with initial conditions $f(0)=1$ has a unique solution.
This can be seen by taking $f,f_0$ two solutions and looking at $g=\frac{f}{f_0}$. You'll get that $$g'=\frac{ff_0^\prime-f'f_0}{f_0^2}$$
But then $$g'=\frac{1}{f_0^2} \left(f \frac{\alpha}{1+x}f_0-\frac{\alpha}{1+x}f f_0\right)=0$$
It follows $g$ is contant, and since $f(0)=f_0(0)=1$, $g=1$ so $f_0=f$. Thus, all you need to show is that $$f=\sum_{n=0}^\infty {\alpha\choose n}x^n$$ converges over $|x|<1$,that so does its derivative and that $(1+x)f'=\alpha f$. This last equality follows from $$n{\alpha\choose n}=\alpha{{\alpha-1}\choose n-1}$$
A much more tedious alternative is possible: you can show that the remainder goes to $0$ for $|x|<1$. To this end, first consdier $0\leq x <1$. Then $$R_{m,0}(x)=\frac{f^{m+1}(t)}{(m+1)!}x^{m+1}$$ for some $t\in(0,1)$. But if $f=(1+x)^{\alpha}$ then $$f^{(m+1)}=\prod_{k=0}^m (\alpha-k)(1+x)^{\alpha-m-1}$$
so that $$R_{m,0}(x)={\alpha\choose m+1}(1+t)^{\alpha-m-1}x^{m+1}$$
But $$0<(1+t)^{\alpha-m-1}\leq 1$$ for $\alpha<m+1$ It follows $\lim_{m\to\infty} R_{m,0}=0$ for any $x\in[0,1)$. Now suppose $-1<x<0$. By Cauchy's remainder formula. $$R_{m,0}=\frac{f^{(m+1)}(t)}{m!}(x-t)^m x$$ for some $x<t\leq 0$.
Yet again $$f^{(m+1)}=\prod_{k=0}^m (\alpha-k)(1+x)^{\alpha-m-1}$$
 so that $$R_{m,0}(x)=(m+1){\alpha\choose m+1}\left(\frac{x-t}{1+t}\right)^m(1+t)^{\alpha-1}x$$
But $$\left| {\frac{{x - t}}{{1 + t}}} \right| = \left| x \right|\left| {\frac{{1 - t/x}}{{1 + t}}} \right|$$ and since $-1<x<t< 0$, $0\leq t/x$ and $0\leq t+1< 1$ so $$\frac{1}{1+t}\geq 1$$ and $$\left| x \right|\left| {\frac{{1 - t/x}}{{1 + t}}} \right| \leqslant \left| x \right|$$
On the other hand, $$|x(1+t)^{\alpha-1}|\leq M_{\alpha,x}$$ where $M_{\alpha,x}=\max(1,(1+x)^{\alpha-1})$. Indeed, if $\alpha \geq 1$ then $0<1+x<1+t<1$ so $0<(1+x)^{\alpha-1}<(1+t)^{\alpha-1}<1$ and if $\alpha<1$ then $1< (1+t)^{\alpha-1}<(1+x)^{\alpha-1}$
It follows that $$\displaylines{
  \left| {{R_{m,0}}\left( x \right)} \right| = \left( {m + 1} \right){\alpha\choose m+1}{\left| {\frac{{x - t}}{{1 + t}}} \right|^m}\left| {{{\left( {1 + t} \right)}^{\alpha  - 1}}x} \right| \cr 
   \leqslant \left( {m + 1} \right){\alpha\choose m+1}{\left| x \right|^{m + 1}}{M_{\alpha ,x}} \cr 
   = \alpha {\alpha-1\choose m}{\left| x \right|^m}{M_{\alpha ,x}} \cr} $$
and this last quantity goes to zero when $m\to\infty$.
All in all $${\left( {1 + x} \right)^\alpha } = \sum\limits_{n = 0}^\infty{\alpha\choose m}  {{x^n}} $$ over $|x|<1$.
