Show that $\sum_{r-k=0}^m\left(\sum_{k=0}^n\binom nk\binom{m}{r-k}x^r\right)=\sum_{r=0}^{m+n}\left(\sum_{k=0}^r\binom nk\binom{m}{r-k}\right)x^r$ 
How does this hold?
  $$\sum_{r-k=0}^m\left(\sum_{k=0}^n\binom nk\binom{m}{r-k}x^r\right)=\sum_{r=0}^{m+n}\left(\sum_{k=0}^r\binom nk\binom{m}{r-k}\right)x^r$$

 A: Consider the relation to be proved, namely,
\begin{align}
\sum_{r=0}^{m+n} \left( \sum_{k=0}^{r} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r} = \sum_{k=0}^{n} \left( \sum_{r=k}^{m} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r}.
\end{align}
Now consider just the left hand side, labeled $S_{L}$,
\begin{align}
S_{L} &= \sum_{r=0}^{m+n} \left( \sum_{k=0}^{r} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r} \\
&= \sum_{r=0}^{m+n} \sum_{k=0}^{m+n} \binom{n}{k} \binom{m}{r} \ x^{r+k} \\
&= \left(\sum_{r=0}^{m+n} \binom{m}{r} \ x^{r} \right) \left(\sum_{r=0}^{m+n} \binom{m}{r} \ x^{k}  \right) \\
&= (1+x)^{m} \  (1+x)^{n}  = (1+x)^{m+n}.
\end{align}
Now consider the right-hand side, labeled $S_{R}$,
\begin{align}
S_{R} &= \sum_{k=0}^{n} \left( \sum_{r=k}^{m} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r} \\
&= \sum_{k=0}^{n} \sum_{r=0}^{m-k} \binom{n}{k} \binom{m}{r} \ x^{r+k} \\
&= \sum_{k=0}^{n} \binom{n}{k} \ x^{k} \left( \sum_{r=0}^{m-k} \binom{m}{r} x^{r} + \sum_{r=m-k+1}^{m+n} \binom{m}{r} x^{r} - \sum_{r=m-k+1}^{m+n} \binom{m}{r} x^{r} \right) \\
&= \left( \sum_{k=0}^{n} \binom{n}{k} \ x^{k} \right) \left( \sum_{r=0}^{m} \binom{m}{r} \ x^{r} \right) \\
&= (1+x)^{m+n}.
\end{align}
Since $S_{L} = S_{R}$ then the identity is shown to be true. 
