MGF of The Negative Binomial Distribution 
For any $0<p<1$ and $r$ a positive integer, the probability function
  $$f(x)={{r+x-1}\choose{x}}p^r(1-p)^x \ \ \ \ \ \ x=0,1,2...$$
  defines a random variable $X$.
  Compute the mgf of $X$ to show that 
  $$m_X(u)=\Big(\frac{p}{1-(1-p)e^u)}\Big)^r \ \ \ \ \ \ \ u<\text{ln}((1-p)^{-1})$$

I have seen many solutions online, but I am still a bit unsure of how to proceed. So far I have,
$$m_X(u)=\sum_{x=1}^{\infty} e^{ux}fx$$
$$m_X(u)=\sum_{x=1}^{\infty} e^{ux}{{r+x-1}\choose{x}}p^r(1-p)^x$$
 I am unsure of how to proceed. I have tried to simplify the above expression to
$${{r+x-1}\choose{x}}=\frac{(r+y-1)!}{(r-1)! \ y!}$$
But this did not yield anything promising.
I saw a solution on this site which used the identity 
$${{-r}\choose{y}}=(-1)^y\frac{(r+y-1)!}{(r-1)! \ y!}$$
But I don't understand where this result has come from, nor how to prove it.
 A: Let's do it your way, and then let's do it another way that may or may not be preferable.  Using your notation,
$$\begin{align*}
m_X(u) &= \sum_{x=0}^\infty e^{ux} \binom{r+x-1}{x} p^r (1-p)^x \\
&= \sum_{x=0}^\infty \binom{r+x-1}{x} p^r ((1-p)e^u)^x \\
&= \frac{p^r}{(1-(1-p)e^u)^r} \sum_{x=0}^\infty \binom{r+x-1}{x} (1 - (1-p)e^u)^r ((1-p)e^u)^x \\
&= \left(\frac{p}{1-(1-p)e^u}\right)^r. \end{align*}$$
Note that the lower index of summation should begin at $x = 0$ since the support of $X$ is $\{0, 1, 2, \ldots\}$. In the third step, I have pulled out a factor of $p^r$, and inserted a factor of $(1 - (1-p)e^u)^r$, neither of which depends on the variable of summation $x$.  Why did I choose this factor to insert?  The reason is that if we recall the PMF of a negative binomial distribution, $$\Pr[X = x] = \binom{r+x-1}{x} p^r (1-p)^x,$$ the relationship between the factors $p^r$ and $(1-p)^x$ are such that the bases must add to $1$.  So long as $$0 < (1-p)e^u < 1,$$ we can think of this as a Bernoulli probability of a single trial; i.e., let $1-p^* = (1-p)e^u$, where $p^*$ is some "modified" Bernoulli probability of some other negative binomial random variable.  In this way, we obtain the sum of probabilities of this "new" negative binomial random variable with parameters $p^*$ and $r$, and the sum of its probabilities over its support is also $1$.

What is the other method?  Well, recall that a negative binomial random variable is simply the sum of $r$ independent and identically distributed geometric random variables; i.e., $$X = Y_1 + Y_2 + \cdots + Y_r,$$ where $Y \sim \operatorname{Geometric}(p)$, with PMF $$\Pr[Y = y] = p(1-p)^y, \quad y = 0, 1, 2, \ldots.$$  Also recall that the MGF of the sum of $r$ iid random variables is simply the MGF of one such random variable raised to the $r^{\rm th}$ power; i.e., $$m_X(u) = \left(m_Y(u)\right)^r.$$  Now if you already know that the MGF of the geometric distribution is $$m_Y(u) = \frac{p}{1-(1-p)e^u},$$ the result immediately follows.  If you don't know this in advance, then you can derive it readily as follows:  $$\begin{align*}
m_Y(u) &= \sum_{y=0}^\infty e^{uy} p (1-p)^y \\
&= p \sum_{y=0}^\infty ((1-p)e^u)^y \\
&= p \cdot \frac{1}{1-(1-p)e^u}, 
\end{align*}$$
where the last step is the consequence of the fact that the sum is an infinite geometric series with common ratio $(1-p)e^u$.  Of course, this imposes the condition $|(1-p)e^u| < 1$, otherwise the series fails to converge.  Since $e^u > 0$ for all $u$, and by construction $0 < p < 1$, it follows that $m_Y(u)$ is defined if and only if $u < -\log(1-p)$.  And this restriction carries over into the MGF of the negative binomial distribution.
