Show that $E[\sum_{i=1}^{N}X_i] =E(N)E(X_1)$ 
Let $(X_n)$ be a sequence of random variables that are independent and identically distributed, with $EX_n < \infty$ and $EX_n^2 < \infty$. Let N be a random variable with range $R \subseteq \Bbb{N}$, independent of $(X_n)$, with $EN < \infty$ and $EN^2 < \infty$. Show that $E[\sum_{i=1}^{N}X_i] =E(N)E(X_1)$

So, I tried to calculate $E(\sum_{i=1}^{N}X_i|N)$. I think that it might be $E(X_1)N$, so, I tried to show that $E[t(N)\sum_{i=1}^{N}X_i]=E[NE(X_1)t(N)]$ for all t which are bounded and Borel-measurable, but I don't know how to calculate $E[t(N)\sum_{i=1}^{N}X_i]$. Any hint?

$\begin{align}
\mathsf E(\sum_{n=1}^N X_n) & = \mathsf E(\mathsf E(\sum_{n=1}^N X_n\mid N)) & \textsf{Iterated Expectation}
\\[1ex] & = ~ & \textsf{Linearity of Expectation}
\\[1ex] & = ~ & \textsf{Independence of $\{X_n\}$ from $N$}
\\[1ex] & = ~ & \textsf{Identical Distribution of $\{X_n\}$}
\\[1ex] & = ~ & \textsf{Series of a Constant}
\\[1ex] & = \mathsf E(N)\;\mathsf E(X_1) & \textsf{Linearity of Expectation}
\end{align}$
Fill in the blanks.
