PDF of the r-th Order Statistic (essentially need help with a derivative or combinatorics) So I get why it is that the CDF of the $r$th order statistic is 
$$ F_{X_{(r)}}(x) = \sum_{i=r}^{n} \binom{n}{i}[F_{X}(x)]^{i}[1-F_{X}(x)]^{n-i} $$
but I'm not seeing how to prove that the PDF is 
$$ f_{X_{(r)}}(t)=\frac{n!}{(r-1)!(n-r)!} [F_{X}(x)]^{r-1}[1-F_{X}(x)]^{n-r}f_{X}(x) $$
I've tried following the explanation given here but I don't understand the jump about half-way down the proof, where it says "Once we do that, we see that the p.d.f. of the $r$th order statistic $Y_r$ is just the first term in the summation in $g_r(y)$." I thought $f_{X_{(r)}}(t) = F'_{X_{(r)}}(t)$ so why would you only select the first term?
 A: The derivative using the chain and product rules is
$$
f_{X_{(r)}}(x) = \sum_{i=r}^{n} \binom{n}{i}\bigg(i[F_{X}(x)]^{i-1}[1-F_{X}(x)]^{n-i}-  (n-i)[F_{X}(x)]^{i}[1 - F_{X}(x)]^{n-i-1} \bigg)f_{X}(x)
$$
Hopefully that part was clear. Next they split the sum into two parts (the parts that got separated by the product rule.) In other words, write $f_{X_{(r)}}(x) = f_{1}(x) + f_{2}(x)$ where
\begin{align*}
f_{1}(x) &=  \sum_{i=r}^{n} \binom{n}{i} i [F_{X}(x)]^{i-1}[1-F_{X}(x)]^{n-i} f_{X}(x) \\
f_{2}(x)&=  -\sum_{i=r}^{n} \binom{n}{i} (n-i) [F_{X}(x)]^{i}[1-F_{X}(x)]^{n-i-1} f_{X}(x) 
\end{align*}
Now they expand out the binomial coefficients,
\begin{align*}
&\binom{n}{i}i = \frac{n!}{i!(n-i)!}i=\frac{n!}{(i-1)!(n-i)!} \\
&\binom{n}{i}(i-n) = \frac{n!}{i!(n-i)!}(n-i) = \frac{n!}{i!(n- i - 1)!}
\end{align*}
Therefore,
\begin{align*}
f_{1}(x) &=  \sum_{i=r}^{n} \frac{n!}{(i-1)!(n-i)!} [F_{X}(x)]^{i-1}[1-F_{X}(x)]^{n-i} f_{X}(x) \\
f_{2}(x)&=  -\sum_{i=r}^{n}  \frac{n!}{i!(n- i - 1)!} [F_{X}(x)]^{i}[1-F_{X}(x)]^{n-i-1} f_{X}(x) 
\end{align*}
Now, note that in $f_{2}$, when $i = n$ the factorial is undefined, so the last term in that sum is actually $n - 1$. This comes from the fact that we absorbed a factor of $(n - i)$ into the factorial, which is $0$ when $i = n$. Let us also extract the $i = r$ term (aka: the first term) out of $f_{1}$. When we do that, the upper limit of the sum will go to $n - 1$ because we got rid of a term, and all the $i$ terms in the summand will go to $i + 1$ because we need to start the remainder of the sum one term later. Thus,
\begin{align*}
f_{1}(x) &= \frac{n!}{(r-1)!(n-r)!} [F_{X}(x)]^{r-1}[1-F_{X}(x)]^{n-r} f_{X}(x) \\
&+\sum_{i=r}^{n - 1} \frac{n!}{i!(n-i - 1)!} [F_{X}(x)]^{i}[1-F_{X}(x)]^{n-i - 1} f_{X}(x) \\
f_{2}(x)&=  -\sum_{i=r}^{n-1}  \frac{n!}{i!(n- i - 1)!} [F_{X}(x)]^{i}[1-F_{X}(x)]^{n-i-1} f_{X}(x) 
\end{align*}
Clearly, $f_{2}$ will cancel out the second term in $f_{1}$ so we're left with
$$
f_{X_{(r)}}(x) = \frac{n!}{(r-1)!(n-r)!} [F_{X}(x)]^{r-1}[1-F_{X}(x)]^{n-r} f_{X}(x)
$$
as required.
