Let $X_1,X_2,...$ be independent random variables each with expectation $\mu$ and $N$ a random variable that.. Let $X_1,X_2,...$ be independent random variables each with expectation $\mu$ and $N$ a random variable that..... takes the value of the set of natural numbers. $N$ and $X_i$ are independent random variables $\forall i=1,2...$. Prove that: 
$$E(\sum_{i=1}^{N}X_i)=\mu EN$$


*

*I will put a star above the equalities that are problematic 


$$E(\sum_{i=1}^{N}X_i)=EE(\sum_{i=1}^{N}X_i|N)=^{\{*1\}}\sum_{n=1}^{\infty}E(\sum_{i=1}^{N}X_i|N=n)P\{N=n\}=\sum_{n=1}^{\infty}E(\sum_{i=1}^{n}X_i|N=n)P\{N=n\}=^{\{*2\}}\sum_{n=1}^{\infty}E(\sum_{i=1}^{n}X_i)P\{N=n\}=...$$ the rest is easy, Im not too sure about these two inequalities. Thoughts?

First off, you don't need to write the expression right before $\{*1\}$. So directly from the first expression to the expression immediately after $\{*1\}$, this is due to conditional expectation.
The equality $\{*2\}$ is true because $\sum_{i=1}^{n}X_i$ is independent of $N$.
