To prove $\sum_{n=0}^\infty \binom{r}{x}\binom{N-r}{n-x}=\binom{N}{n}.$ To prove $$\sum_{x=0}^n \binom{r}{x}\cdot \binom{N-r}{n-x}=\binom{N}{n}.$$
I tried comparing the coefficients of
$(1+x)^{(n+k)} = (1+x)^n(1+x)^k$
but couldn't reach the answer. 


We  obtain for $0\leq   r\leq N$
  \begin{align*}
(1+x)^{N}&=(1+x)^r(1+x)^{N-r}\\
&=\sum_{k=0}^r\binom{r}{k}x^k\sum_{l=0}^{N-r}\binom{N-r}{l}x^l\\
&=\sum_{n=0}^{N}\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\binom{r}{k}\binom{N-r}{l}\right)x^n\\
&=\sum_{n=0}^{N}\left(\sum_{k=0}^n\binom{r}{k}\binom{N-r}{n-k}\right)x^n\\
\end{align*}
  Since
  \begin{align*}
(1+x)^{N}=\sum_{n=0}^{N}\binom{N}{n}x^n
\end{align*}
  we compare coefficients and conclude
  \begin{align*}
\sum_{k=0}^n\binom{r}{k}\binom{N-r}{n-k}=\binom{N}{n}
\end{align*}

