Proving the equation using binomial theorem I want to prove this theorem using Binomial theorem and I've got trouble in understanding 3rd step if anyone knows why please explain :)                         Prove that sum:     
$\sum_{r=0}^{k}\binom{m}{r}\binom{n}{k-r}=\binom{m+n}{k}$ 
1st step:
$(1+y)^{m+n}=(1+y)^m(1+y)^n$ 
2nd step: we use the binomial theorem formula and evaluate that sum 
$\sum_{m+n}^{r=0}\binom{m+n}{r}y^r=\sum_{r=0}^{m}\binom{m}{r}y^r \sum_{r=0}^{n}\binom{n}{r}y^r$
3step: equating the coefficient of $y^k$ 
$\binom{m+n}{k}=\binom{m}{0}\binom{n}{k}+\binom{m}{1}\binom{n}{k-1}+...+\binom{m}{k}\binom{n}{k-k}$
Why and how is the 3rd step allowed? Thank you             

By the equation of 2nd step,
Left:
$$\binom{m+n}{0}y^{0}+\binom{m+n}{1}y^{1}+\cdots+\binom{m+n}{m+n-1}y^{m+n-1}+\binom{m+n}{m+n}y^{m+n}$$
The coefficient of $y^{k}$ is $\binom{m+n}{k}$
Right:
$$(\binom{m}{0}y^{0}+\cdots+\binom{m}{m}y^{m})(\binom{n}{0}y^{0}+\cdots+\binom{n}{n}y^{n}) $$
The term $y^{k}$ come from 
$$\binom{m}{0}y^{0}\binom{n}{k}y^{k}+\binom{m}{1}y^{1}\binom{n}{k-1}y^{k-1}+ \cdots +\binom{m}{k-1}y^{k-1}\binom{n}{1}y^{1}+\binom{m}{k}y^{k}\binom{n}{0}y^{0}$$
So the coefficient of $y^{k}$ is $\binom{m}{0}\binom{n}{k} + \binom{m}{1}\binom{n}{k-1}+ \cdots +\binom{m}{k-1}\binom{n}{1}+\binom{m}{k}\binom{n}{0}$
By the equality of the equation of 2nd step, the coefficient of $y^{k}$ of left and right might be the same, so $$\binom{m+n}{k} = \binom{m}{0}\binom{n}{k} + \binom{m}{1}\binom{n}{k-1}+ \cdots +\binom{m}{k-1}\binom{n}{1}+\binom{m}{k}\binom{n}{0}$$
