Stuck on a step for finding the closed formula for catalan numbers from generating function! I am looking at the Frazer Jarvis paper, entitled Catalan Numbers.
http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec01.pdf
In this paper, he derives the closed formula from the generating function.
The $nth$ Catalan number, denoted $C_n$, is the number of ways
of multiplying together $n$ symbols.
I am stuck on page 7, particularly the following steps:
$\begin{equation}C_n=-\frac{1}{2}\bigg\{\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2n-3}{2})}{n!}(-4)^n\bigg\}\end{equation}$
$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}$
$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot 1 \cdot\frac{3}{2} \cdot 2...(n-2)(\frac{2n-3}{2})(n-1)]}{n!(n-1)!}2^{2n}\bigg\}$
$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(1 \cdot 2 \cdot 3 \cdot 4...(2n-4)(2n-3)(2n-2)}{n!(n-1)!}2^{2}\bigg\}$
Particularly, from the 2nd to the 4th line, I am struggling to understand the steps taken here.
I am confused with what he is doing with the $\frac{1}{2}$. Is he pulling it out in the 3rd step? And most of all, I am very confused with how the $2^n$ went away in the 4th step. If anyone could please help me understand some of the steps that he skipped here, I would really appreciate it! Thank you so much!!!

Step 1 (in the question): $\begin{equation}C_n=-\frac{1}{2}\bigg\{\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2n-3}{2})}{n!}(-4)^n\bigg\}\end{equation}$
Step 2: $C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}$
In step 1, $-\frac{1}{2}, -\frac{3}{2}... -\frac{2n-3}{2}$ has $(n-1)$ terms with negative signs as $2n - 3 = 1 + 2 (n-2)$ and there is $-\frac{1}{2}$ outside curly brackets too. So the negative sign is taken out in step 2 as $(-1)^n \times (-1)^n = (-1)^{2n} = 1$.
Also note there are $n$ $2's$ in the denominator of $\frac{1}{2} (-\frac{1}{2}), (-\frac{3}{2})... (-\frac{2n-3}{2})$
Step 3: $C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot 1 \cdot\frac{3}{2} \cdot 2...(n-2)(\frac{2n-3}{2})(n-1)]}{n!(n-1)!}2^{2n}\bigg\}$
i) Writing $4^n$ as $2^{2n}$, ii) In step 2, we multiplied numerator and denominator by $(n-1)!$. We now expand $(n-1)!$ in the numerator as $1 \times 2 \times 3 ...\times (n-1)$ and write in order.
Step 4: $C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(1 \cdot 2 \cdot 3 \cdot 4...(2n-4)(2n-3)(2n-2)}{n!(n-1)!}2^{2}\bigg\}$
In this step we cancel $2^{2n-2}$ in the numerator with $2^{n-1}$ in denominator and we also use $2^{n-1}$ to double each term of $(n-1)!$ from $1 \times 2 \times 3 ..(n-1)$ to $2 \times 4 \times 6 ... \times (2n-2)$ which will lead to numerator as $(2n-2)!$
