Show that $K^{(m)}(1)$ is equal to the $m$th factorial moment. 
$\textbf{1.98.}$  Let $X$ be a random variable such that $K(t)=E(t^X)$ exists for all real values of $t$ in a certain open interval that includes the point $t=1$. Show that $K^{(m)}(1)$ is equal to the $m$th factorial moment $E[X(X-1)\cdots(X-m+1)]$.


Let's first show that for $n\in \mathbb{N}$, if $v(t)=t^x$ and $v^{(n)}(t)$ means the $n$th derivative of $v$ with respect to $t$, then $v^{(n)}(t)=x(x-1)\cdots(x-n+1)t^{x-n}$. We will show this by induction.
Base case: $n=1$
\begin{gather}
v^{(1)}(t)=xt^{x-1}, \text{ by the power rule since } x \text{ is held constant.}\tag1\\
\text{Letting } n=1 \text{ in the equation } x(x-1)\cdots(x-n+1)t^{x-n} \text{ provides } xt^{x-1}\tag2\\
\text{Thus } v^{(n)}(t)=x(x-1)\cdots(x-n+1)t^{x-n} \text{ for } n=1, \text{ closing the base case.}\tag3
\end{gather}
Inductive step
\begin{gather}
\text{ Suppose when } n=m, \text{ we have that } v^{(m)}(t)=x(x-1)\cdots(x-m+1)t^{x-m} \text{ holds true}.\tag4\\
\text{ Consider when } n=m+1. \text{ Then } v^{(m+1)}(t)=\frac{d}{dt}\left[v^{(m)}(t)\right]=x(x-1)\cdots(x-m+1)(x-m)t^{x-(m+1)}\tag5\\
x(x-1)\cdots(x-n+1)t^{x-n} \text{ evaluated at } n=m+1 \text{ provides }x(x-1)\cdots(x-m+1)(x-m)t^{x-(m+1)}\tag6\\
\text{ Thus we must have } v^{(n)}(t)=x(x-1)\cdots(x-n+1)t^{x-n}, \text{ closing the induction}.\tag7
\end{gather}

Now that we got that out of the way, let us consider when random variable $X$ is continuous with p.d.f. $f(x)$. Let $K^{(m)}(t)$ represent the $m$th derivative of $K$ with respect to $t$. Then
\begin{align}
K(t)=E(t^X)=E[v(t)]=\int_{-\infty}^{\infty}v(t)f(x)\ dx\tag8\\
K^{(m)}(t)=\int_{-\infty}^{\infty}v^{(m)}(t)f(x)\ dx\tag9 
\end{align}
I'm justifying $(9)$ by some theorem in analysis that allows us to change the order of differentiation and integration, but I will not state it here because...please don't ask me to state it here. Anyway, since we showed $v^{(n)}(t)=x(x-1)\cdots(x-n+1)t^{x-n}$, we have
\begin{align}
K^{(m)}(t)&=\int_{-\infty}^{\infty}x(x-1)\cdots(x-m+1)t^{x-m}f(x)\ dx\tag{10}\\
K^{(m)}(1)&=\int_{-\infty}^{\infty}x(x-1)\cdots(x-m+1)f(x)\ dx\tag{11}\\
&=E[X(X-1)\cdots(X-m+1)]\tag{12}
\end{align}
Similar proof for the discrete case.

Have I proven this correctly? Is there a faster way of doing so? Any feedback is greatly appreciated!
 A: You have correctly proved that $\frac{d^n}{dt^n}t^x=x(x-1)\cdots(x-n+1)t^{t-n}$.
The rest of your solution is correct (except for the omitted justification of $(9)$), but only applies to the case when $X$ has a pdf. What about when $X$ is discrete, or has the Cantor distribution? The proof which unites all of these cases is as simple as
$$
K^{(m)}(1)=\frac{d^m}{dt^m }E[t^X]\Big|_{t=1} \stackrel{?}=E\left[\frac{d^m}{dt^m }t^X\Big|_{t=1}\right]=E[X(X-1)\cdots (X-n+1)\cdot 1^{X-n}],
$$
though we still need to justify the $\stackrel?=$ step. This needs to be proved in $m$ stages, where during the $i^{th}$ stage, we move the $i^{th}$ derivative into the expectation, as follows:
$$
\frac{d^{m-i+1}}{dt^{m-i+1}}E\left[\frac{d^{i-1}}{dt^{i-1}}t^X\right]=
\frac{d^{m-i}}{dt^{m-i}}E\left[\frac{d^{i}}{dt^{i}}t^X\right]\tag{$*$}
$$
To justify changing the order of $\frac{d}{dt}$ and $E$, I will use this theorem from Folland:

2.27 Theorem Let $(X,\mathcal M, \mu)$ be a measure space,  let $f:X\times [a,b]\to\mathbb C$ be a measurable function such that $f(\cdot, t)$ is integrable for each $t\in [a,b]$, and let $F(t)=\int f(x,t)\,\mu(dx)$. If $\partial f/\partial t$ exists, and there is a $g\in L^1(\mu)$ such that $|\partial f/\partial t(x,t)|\le g(x)$ for all $(x,t)\in X\times [a,b]$, then $F$ is differentiable and $F'(x)=\int \partial f/\partial t(x,t)\,\mu(dx)$.

Onto the problem at hand, suppose that $K(t)$ exists for all $t\in [a_0,b_0]$, where $0<a_0<1<b_0$. Choose $a,b$ so that $a_0<a<1<b<b_0$. For each $i\in \{1,\dots,m\}$, in order to apply Theorem 2.27 to $(*)$, we need to find a dominating random variable $Y_i$ so that $E|Y_i|<\infty$, and
$$
\left|\frac{d^i}{dt^i}t^X\right|=\left|X(X-1)\cdots (X-i+1) t^{X-i}\right|\le Y_i
$$
for all values of $X$, and all $t\in [a,b]$. Let $$P_i(x)=\frac{x(x-1)\cdots(x-i+1)}{t^i},$$so that $\frac{d^i}{dt^i}t^X=P_i(X)t^X$. The only fact about $P_i(x)$ we need to pay attention to is that it is a polynomial. We then bound $|P_i(X)t^X|$ as follows:
\begin{align}
|P_i(X)t^X|
  &=|P_i(X)|t^X\cdot 1_{X\ge 0}
    &&+&&
    |P_i(X)|(1/t)^{-X}\cdot 1_{X< 0}
\\&=|P_i(X)|(t-b_0)^X\cdot b_0^X\cdot 1_{X\ge 0}
    &&+&&
    |P_i(X)|(1/t-1/a_0)^{-X}\cdot a_0^X\cdot 1_{X< 0}
\\&\le \color{blue}{|P_i(X)|(b-b_0)^X\cdot 1_{X\ge 0}}\cdot b_0^X
    &&+&&
    \color{#F60}{|P_i(X)|(1/a-1/a_0)^{-X}\cdot 1_{X< 0}}\cdot a_0^X
\\&\le \color{blue}{C_1} b_0^X &&+&& \color{#F60}{C_2}a_0^X
\end{align}
Here, $C_1$ and $C_2$ are certain constants which do not depend on $\omega$ or $t$ (only on $a_0,a,b,b_0$ and $i$). These exist since both $|P_i(X)|(b-b_0)^X\cdot 1_{X\ge 0}$ and $|P_i(X)|(1/a-1/a_0)^{-X}\cdot 1_{X< 0}$ are a polynomial in $X$ times a decaying exponential, so the result approaches zero as $X\to\infty$ and is therefore bounded.
We have proven that $|P_i(t)t^X|\le C_1b_0^X+C_2a_0^X$. Since both $a_0^X$ and $b_0^X$ are integrable by assumption, we have found our dominating function.
