TITLE: Binomial-Theorem proof
QUESTION [5 upvotes]: I have the following problem:
$n,m \in \mathbb{N_0}$. Write $(1+x)^{n+m}=(1+x)^{n}(1+x)^{m}$ with the help of binomial formulas, multiply the right side, and deduce the following
Identity between binomial coefficients:
$\begin{pmatrix} n+m\\k\\ \end{pmatrix}= \sum_{l=0}^{k}{\begin{pmatrix} n\\l\\ \end{pmatrix}}{\begin{pmatrix} m\\k-l\\ \end{pmatrix}}  \forall k \in \mathbb{N_0}$
My idea:
I know that: $(a+b)^n=\sum_{k=0}^{n}{\begin{pmatrix} n\\k\\ \end{pmatrix}a^{n-k}b^k=\sum_{k=0}^{n}{\begin{pmatrix} n\\k\\ \end{pmatrix}a^{k}b^{n-k}}}$
So:$(1+x)^{n+m}=\sum_{k=0}^{n}{{\begin{pmatrix} n\\k\\ \end{pmatrix}x^k}*\sum_{k=0}^{n}{{\begin{pmatrix} m\\k\\ \end{pmatrix}x^k}}} =\sum_{k=0}^{n}{{\begin{pmatrix} n+m\\k\\ \end{pmatrix}x^k}}=(1+x)^{n+m}$
But this doens't seem right.. Can anyone help me here?

REPLY [5 votes]: We obtain
  \begin{align*}
\sum_{k=0}^{n+m}\color{blue}{\binom{n+m}{k}}x^k
&=(1+x)^{n+m}=(1+x)^n(1+x)^m\\
&=\sum_{l=0}^n\binom{n}{l}x^l\sum_{j=0}^m\binom{m}{j}x^j\\
&=\sum_{k\geq 0}\left(\sum_{{l+j=k}\atop{k,j\geq 0}}\binom{n}{l}\binom{m}{j}\right)x^k\tag{1}\\
&=\sum_{k\geq 0}\left(\color{blue}{\sum_{l=0}^k\binom{n}{l}\binom{m}{k-l}}\right)x^k\tag{2}\\
\end{align*}

Comment:

In (1) we  apply  the Cauchy product formula. Observe the series starting with index $k=0$ is finite since $\binom{r}{k}=0$ if $k>r$, $r\in\mathbb{N}$. 
In  (2) we  set the  upper limit of the inner series to $k$ since $\binom{m}{k-l}=0$ if $l\geq k$.