<s>[INST] How to express $\binom{a+b}{n}$ as a sum of regular coefficients I am trying to prove that $(1 + f(x))^a(1 + f(x))^b = (1 + f(x))^{a+b}$ in the world of formal power series. At a certain point in the prove I get
\begin{align*}
  (1 + f(x))^a(1+f(x))^b& = \ldots \\
  & = \lim\limits_{N \to \infty} \left(\left[\sum_{n=0}^{N}\binom{a}{n} f(x)^n\right] \cdot \left[\sum_{n=0}^{N}\binom{b}{n} f(x)^n\right]\right) \\
  & = \lim\limits_{N \to \infty} \left(\left[\binom{a}{0} \binom{b}{0} \right] f(x)^0 + \left[\binom{a}{1} \binom{b}{0} + \binom{a}{0} \binom{b}{1}\right] f(x)^1 + \ldots \right) \\
  & = \lim\limits_{N \to \infty} \sum_{n=0}^{N} \binom{a+b}{n} f(x)^n 
\end{align*}
I tried $\binom{a}{0} \binom{b}{0} = \frac{1}{0!} \cdot \frac{1}{0!} = \frac{(a+b)^{\underline{0}}}{0!}$, $\binom{a}{1} \binom{b}{0} + \binom{a}{0} \binom{b}{1} = \frac{a}{1!} \cdot \frac{1}{0!} + \frac{1}{0!} \frac{b}{1!} = \frac{(a+b)^{\underline{1}}}{1!}$, $\ldots$, but I don't seem to be able to generalize and prove this.
Could anybody get me in the right direction of how to prove that $\left[\sum_{n=0}^{N}\binom{a}{n} f(x)^n\right] \cdot \left[\sum_{n=0}^{N}\binom{b}{n} f(x)^n\right] = \sum_{n=0}^{N} \binom{a+b}{n} f(x)^n$?
Or am I approaching this problem the wrong way?
 [/INST] The development chain goes like this:
$$
\eqalign{
  & \left( {1 + f(x)} \right)^a \left( {1 + f(x)} \right)^b  = \sum\limits_k {\left( \matrix{
  a \cr 
  k \cr}  \right)f(x)^k } \sum\limits_j {\left( \matrix{
  b \cr 
  j \cr}  \right)f(x)^j }   \cr 
  &  = \sum\limits_j {\sum\limits_k {\left( \matrix{
  a \cr 
  k \cr}  \right)\left( \matrix{
  b \cr 
  j \cr}  \right)f(x)^k } } f(x)^j  =   \cr 
  &  = \sum\limits_n {\left( {\sum\limits_k {\left( \matrix{
  a \cr 
  k \cr}  \right)\left( \matrix{
  b \cr 
  n - k \cr}  \right)} } \right)\;} f(x)^n  =  \cr} 
$$
here the Chu-Vandermonde identity comes into play,(https://en.wikipedia.org/wiki/Vandermonde's_identity) giving:
$$
\eqalign{
  &  = \sum\limits_n {\left( \matrix{
  a + b \cr 
  n \cr}  \right)\;} f(x)^n  =   \cr 
  & \left( {1 + f(x)} \right)^{a + b}  \cr} 
$$
</s>