Exchange the order of summations: $\sum_{k=0}^{n}\sum_{m=k}^{n}\binom{m}{k}e_mz^k=\sum_{m=0}^{n}e_m\sum_{k=0}^{m}\binom{m}{k}z^k$ I don't understand how it work from the second line to third line. Is there some rule/technique or it's just by observation? Why it seems that $m$ is depend on $k$ then the third line says they're independent?
The $L$ and $E$ are two generating functions, and $l_k$ and $e_m$ are just constant once $k,m$ are fixed.
\begin{align*}
L(z)&=\sum_{k=0}^{n}l_kz^k\\
&=\sum_{k=0}^{n}\sum_{m=k}^{n}\binom{m}{k}e_mz^k\\
&=\sum_{m=0}^{n}e_m\sum_{k=0}^{m}\binom{m}{k}z^k\\
&=\sum_{m=0}^{n}e_m(z+1)^m\\
&=E(z+1)
\end{align*}
 A: This is in effect asking why
$$\sum_{k=0}^n\sum_{m=k}^n f(k,m)=\sum_{m=0}^n\sum_{k=0}^mf(k,m).$$
This is because both sides are the sum of $f(k,m)$ over all pairs of integers
$(k,m)$ with $0\le k\le m\le n$.
A: This is due to 
$$\{(k, m) : 0 \le k \le n , k \le m \le n \}= \{(k, m): 0 \le m \le n , 0 \le k \le m \}$$
For the set on the left, $k$ appears as the lower bound of $m$ but on the right. We first describe the values that $k$ can take and fix it, and then we describe the values that $m$ can take.
When we switch the order, $m$ is the upper bound of $k$. We first describe the values that $m$ can take and fix it, and then we describe the values that $k$ can take.
A: The reason because the equality
$$
\sum_{k=0}^{n}\sum_{m=k}^{n}\binom{m}{k}e_mz^k=\sum_{m=0}^{n}e_m\sum_{k=0}^{m}\binom{m}{k}z^k
$$
basically holds is simply a matter of reordering of the terms, as Lord Shark the Unknown and Siong Thye Goh noted in their answer: formally, the proof proceeds by noting that, since
$$
(m)_k=\prod_{i=1}^k(m-i+1)=0\quad 0\le m<k\quad k,n\in\mathbb{N}
$$
then
$$
\binom{m}{k}=\frac{(m)_k}{k!}=0\quad\text{ if }\quad k>m 
$$
from the properties of general factorials. Then we have that 
$$
\begin{split}
\sum_{m=0}^{k-1}\binom{m}{k}e_mz^k&=0\quad\text{ if }0\le m< k\\
\sum_{k=m+1}^{n}\!\!\binom{m}{k}e_mz^k&=0\quad\text{ if }0\le k< m\le n\\
\end{split}
$$
(remember that $\sum_{i=p}^{q} a_i=0\text{ if }p>q$) and 
$$
\begin{split}
\sum_{k=0}^{n}\sum_{m=k}^{n}\binom{m}{k}e_mz^k &=\sum_{k=0}^{n}\left[\sum_{m=k}^{n}\binom{m}{k}e_mz^k+\sum_{m=0}^{k-1}\binom{m}{k}e_mz^k\right]\\
&=\sum_{k=0}^{n}\sum_{m=0}^{n}\binom{m}{k}e_mz^k\\
&=\sum_{m=0}^{n}\sum_{k=0}^{n}\binom{m}{k}e_mz^k\\
&=\sum_{m=0}^{n}\left[\sum_{k=0}^{m}\binom{m}{k}e_mz^k+\sum_{k=m+1}^{n}\binom{m}{k}e_mz^k\right]\\
&=\sum_{m=0}^{n}e_m\sum_{k=0}^{m}\binom{m}{k}z^k\\
\end{split}
$$
Note: the process I have shown is effective in showing how to do the reordering when there are some combination of the index $k,m\le n$ for which $f(k,m)=0$. I learned it long ago, in the years 1989-1990 during the course of Mathematical Analysis I at the faculty of engineering of the University of Bologna.
