Moment Generating Functions of Normal/Gaussian Random Variable: 3rd, 4th,..,kth Moment Derivation of Normal Random MGF:

I'm having trouble deriving the answers for $E\,[X^3]$ and $E\,[X^4]$ given the information from the image I've posted. In the image I understand how they setup the equation for normal random distribution $X$: given as $G(\theta)$. 
What I am not understanding is how you can go from simply completing the square to finding the 4th moment. Am I simply misinterpreting the equation given in the image or do I need to complete more steps that are not stated in the image?
 A: You've got that your MGF is 
\begin{equation}
 G(\theta) = \exp(\mu \theta + \frac{\sigma^2\theta^2}{2})
\end{equation}
This means that you can get $E(X^n)$ for any $n \geq 1$, using the following
\begin{equation}
 E(X^n) = \frac{d^n}{d \theta^n} G(\theta) \Big\vert_{\theta=0} =G^{(n)}(0) \tag{1}
\end{equation}
So, let's get the four derivatives
\begin{align}
 G'(\theta) &= (\sigma^2 \theta + \mu)G(\theta)\\
 G''(\theta) &= (\sigma^4 \theta^2 + 2\mu\sigma^2\theta +\sigma^2 + \mu^2)G(\theta)\\
 G'''(\theta) &= (\sigma^2 \theta + \mu)(\sigma^4 \theta^2 + 2\mu\sigma^2\theta +3\sigma^2 + \mu^2)G(\theta)\\
 G''''(\theta) &= (\sigma^8 \theta^4 + 4 \mu \sigma^6 \theta^3 + 6\sigma^4(\sigma^2 + \mu^2)\theta^2 + 4\mu\sigma^2(3\sigma^2 + \mu^2)\theta + 3\sigma^4 + 6\mu^2\sigma^2 + \mu^4)G(\theta)
\end{align}
Using equation $(1)$, we get that
\begin{align}
 E(X) &= G'(0)\\
 E(X^2) &= G''(0)\\
 E(X^3) &= G'''(0)\\
 E(X^4) &= G''''(0)
\end{align}
Replacing
\begin{align}
 E(X) &=  \mu \\
 E(X^2) &= \sigma^2 + \mu^2\\
 E(X^3) &=  \mu(3\sigma^2 + \mu^2)\\
 E(X^4) &= ( 3\sigma^4 + 6\mu^2\sigma^2 + \mu^4)
\end{align}
A: The completing the square in the exponent is to show
$$
G(\theta)=\mathbb{E}e^{\theta X}=\exp\left(\mu\theta+\frac12\sigma^2\theta^2\right)
$$
because the integrand $e^{\theta x}f_X(x)$ is $\exp\left(\mu\theta+\frac12\sigma^2\theta^2\right)$ times the probability density function of a $N(\mu+\sigma^2\theta,\sigma^2)$ distribution.
Now you have $G(\theta)$, which you know is analytic about $\theta=0$, you can differentiate it as many times as you please and find the value of $\mathbb{E}[X^n]=G^{(n)}(0)$
