Sum over product of two binomial distributions 
The problem is that of a two-stage "binomial experiment", where first a number $k$ out of $n$ is drawn (each element with probability $p_1$) and later a number $m$ out of those $k$ is drawn (each element with probability $p_2$). As shown here https://stats.stackexchange.com/questions/68800/stochastic-inequality-with-product-of-binomial-distributions, it turns out that $m$ ~ $Bin(n,p_1p_2)$.

This is shown by this equation/theorem:
\begin{equation}
\sum _{k=m}^n {p_1}^k {p_2}^m \binom{k}{m} \binom{n}{k} (1-{p_2})^{k-m} (1-{p_1})^{n-k} = \binom{n}{m} ({p_1} {p_2})^m (1-{p_1} {p_2})^{n-m}
\end{equation}
Can anybody help me / tell me how to proof this formally? It looks a bit magical for me.
 A: Put $p=p_1, q=p_2$ to avoid messy subscripts. 
Midway, put $r=k-m$ and $N=n-m$ to simplify notation.
$$\begin{align}
&\sum _{k=m}^n \color{green}{{p_1}^k {p_2}^m} \color{orange}{\binom{k}{m} \binom{n}{k}} (1-{p_2})^{k-m} (1-{p_1})^{n-k}\\
&=\sum_{k=m}^{n}\color{green}{p^kq^m}\color{orange}{\binom nk \binom km}(1-q)^{k-m}(1-p)^{n-k}\\
&=\sum_{k=m}^{n}\color{green}{(pq)^mp^{k-m}}\color{orange}{\binom nm \binom {n-m}{k-m}}(1-q)^{k-m}(1-p)^{n-k}\\
&=\color{orange}{\binom nm} \color{green}{(pq)^m}\sum_{k=m}^{n}\color{orange}{\binom {n-m}{k-m}}\color{green}{p^{k-m}}(1-q)^{k-m}(1-p)^{n-k}\\
&=\binom nm (pq)^m\sum_{r=0}^N \binom N{r}\left[p(1-q)\right]^r(1-p)^{N-r}\\
&=\binom nm (pq)^m[p(1-q)+(1-p)]^N\\
&=\binom nm (pq)^m(1-pq)^{n-m}\\
&=\binom{n}{m} ({p_1} {p_2})^m (1-{p_1} {p_2})^{n-m}\qquad \blacksquare
\end{align}$$
A: \begin{equation}
\sum _{k=m}^n {p_1}^k {p_2}^m \binom{k}{m} \binom{n}{k} (1-{p_2})^{k-m} (1-{p_1})^{n-k} =\\
p_1^mp_2^m\sum _{k=m}^n {p_1}^{k-m} \frac{k!}{m!(k-m)!}\frac{n!}{k!(n-k)!} (1-{p_2})^{k-m} (1-{p_1})^{n-k} =\\
p_1^mp_2^m\sum _{k=m}^n \frac{(n-m)!}{(n-k)!(k-m)!}\frac{n!}{m!(n-m)!} (p_1(1-{p_2}))^{k-m} (1-{p_1})^{n-k} =\\
p_1^mp_2^m\sum _{k=m}^n \binom{n-m}{n-k} \binom{n}{m}(p_1(1-{p_2}))^{k-m} (1-{p_1})^{n-k} =\\
p_1^mp_2^m\binom{n}{m}\sum _{k=m}^n \binom{n-m}{n-k} (p_1(1-{p_2}))^{k-m} (1-{p_1})^{n-k} =\\
p_1^mp_2^m\binom{n}{m}(p_1(1-{p_2})+1-{p_1})^{n-m}=\\
\binom{n}{m} ({p_1} {p_2})^m (1-{p_1} {p_2})^{n-m}
\end{equation}
