$\mathbb{E}[e^{Xt}] = \mathbb{E}[\mathbb{E}[e^{Xt}\mid Y]] = \mathbb{E}[M_{X\mid Y}(t)]$? $$\mathbb{E}[e^{Xt}] = \mathbb{E}[\mathbb{E}[e^{Xt}\mid Y]] = \mathbb{E}[M_{X\mid Y}(t)]$$
How do I get the above statement? I don't understand how in the 1st step $e^{Xt}=\mathbb{E}[x^{Xt}\mid Y]$ then in the 2nd $\mathbb{E}[e^{Xt}\mid Y] = M_{X\mid Y}(t)$?

This is from the part (b) of the below question: 

And its provided solution: (see 1st 3 lines)

 A: The first step is what is called the law of Iterated Expecations. Simply put, if $X,Y$ are random variables then, $$\mathbb{E}_X[X] = \mathbb{E}_Y[\mathbb{E}_{X \mid Y}[X \mid Y]]$$ and by definition, $\mathbb{E}_{X \mid Y}[X \mid Y]$ is a function of $Y$ and again by definition $\mathbb{E}_{X \mid Y}[e^{xt}\mid Y]$ is the Moment generating function of the random variable defined by the PDF $f_{X\mid Y}(x\mid y)$.
A: The law of total expectation, or law of iterated expectations, is what is being used here.
A simple example: Suppose the distribution of a pair of random variables $U,V$ is given by the table below.  This means that, for example, $\Pr(U=1\ \&\ V=2)=2/16$, etc. 
$$
\begin{array}{c|ccc}
 & V=1 & V=2 & V=3 \\
\hline
U=1 & 1/16 & 2/16 & 3/16 \\
U=2 & 3/16 & 4/16 & 3/16
\end{array}
$$
Then the conditional distribution of $U$ given the event $V=1$ is given by $\Pr(U=1\mid V=1)=1/4$ and $\Pr(U=2\mid V=1)=3/4$.
The conditional distribution of $U$ given the event $V=2$ is given by $\Pr(U=1\mid V=2)=1/3$ and $\Pr(U=2\mid V=2)=2/3$.
The conidtional distribution of $U$ given the event $V=3$ is given by $\Pr(U=1\mid V=3)=1/2$ and $\Pr(U=2\mid V=3)=1/2$.
So the conitional expected value of $U$ given $V=1$ is $\mathbb E(U\mid V=1)=1.75=\dfrac74$.
The conditional expected value of $U$ given $V=2$ is $\mathbb E(U\mid V=2) = 1.666666\ldots = \dfrac53$.
The conditional expected value of $U$ given $V=3$ is $\mathbb E(U\mid V=3) = 1.5=\dfrac32$.
Now we can say
$$
\mathbb E(U\mid V)=\begin{cases} 1.75=7/4 & \text{if }V=1, \\[8pt]
1.666666\ldots=5/3 & \text{if }V=2, \\[8pt]
1.5=3/2 & \text{if }V=3. \end{cases}
$$
Thus $\mathbb E(U\mid V)$ is a random variable in its own right, equal to either $1/75=7/4$, $1.666666\ldots=5/3$, or $1.5=3/2$, with respective probabilities $4/16$, $6/16$, and $6/16$.  It has its own expected value:
$$
1.75\cdot\frac{4}{16} + (1.666666\ldots)\cdot\frac{6}{16}+1.5\cdot\frac{6}{16}
$$
$$
=\frac74\cdot\frac{4}{16}+\frac53\cdot\frac{6}{16} + \frac32\cdot\frac{6}{16} = \frac{84+120+108}{12\cdot16} = \frac{312}{192} = \frac{13}{8} = 1.625.
$$
That is what is meant by $\mathbb E(\mathbb E(U\mid V))$.
The law of total expecation says that is the same number as $\mathbb E(U)$.
So when you say $\mathbb E(U) = \mathbb E(\mathbb E(U\mid V))$, that is what is meant.
It does not mean that $U$ is equal to $\mathbb E(U\mid V)$.  The fact that two things have the same expected value does not mean they are the same thing.
