Finding the Moment Generating function of a Binomial Distribution Suppose $X$ has a $\rm{Binomial}(n,p)$ distribution. Then its moment generating function is

\begin{align}
M(t)
&= \sum_{x=0}^x e^{xt}{n \choose x}p^x(1-p)^{n-x} \\
&=\sum_{x=0}^{n} {n \choose x}(pe^t)^x(1-p)^{n-x} \\
&=(pe^t+1-p)^n
\end{align}

Can someone please explain how the sum is obtained from lines (2) to (3)?
 A: The moment generating function for the binomial distribution $B_{n,p}$, whose discrete density is $\binom{n}{k}p^k(1-p)^{n-k}$, is defined as
$$
\begin{align}
M_{B_{n,p}}(t)
&=\mathrm{E}(e^{tk})\\
&=\sum_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}e^{tk}\\
&=\sum_{k=0}^n\binom{n}{k}\left(pe^t\right)^k(1-p)^{n-k}\\
&=\left(pe^t+(1-p)\right)^n
\end{align}
$$
The last step is simply an application of the binomial theorem.
A: Adding to @krngrvr09 's response:
Because Bernoulli is a special case of Binomial distribution, PMF of binomial distribution $$\binom{n}{k}p^k(1-p)^{n-k}$$ can be rewritten as $$\binom{1}{0}p^1(1-p)^{1-0}$$
Thus, we can get the following for each instance of $X$ for $X \sim Bin(n,p)$ and $X = \sum^n_{j=1}X_j$
$$ 
\begin{array}
  \mathbb{E}[e^{tX_j}] & = \sum^n_{k=0}e^{tk}\binom{1}{0}p^1(1-p)^{1-0} \\
                      & = (pe^t +1 -p)^1 \ \ \ \ \ \ \ \text{by Binomial Theorem}\\
\end{array}
$$
A: $\phi(t) = \mathbb{E}[e^{tX}]
       \Rightarrow \mathbb{E}[e^{t\cdot(\Sigma x)}]
       \Rightarrow \mathbb{E}[e^{tx_1}\cdot e^{tx_2}\cdot...\cdot e^{tx_n}]
       \Rightarrow \mathbb{E}[e^{tx_1}]\cdot \mathbb{E}[e^{tx_2}]\cdot...\cdot \mathbb{E}[e^{tx_n}]$
Since all individual events are independent,
$\Rightarrow [pe^t + (1-p)].[pe^t + (1-p)].[pe^t + (1-p)]...[pe^t + (1-p)]$ n times, since all n random variables are Bernoulli random variables
$\Rightarrow[pe^t + (1-p)]^n=[pe^t + q]^n$, where $q=1-p$.
