I am wondering if anyone can help to shed some light on something that I think should be very easy but I dont quite understand.
In my textbook, how does author make this conclusion,
From $$ a_n=\frac{(-1)^n}{[3 \times 5 \times 7 .. \times (2n+1)]n!}a_0 $$ for $n \ge 4$
to $$a_n=\frac{(-1)^n2^n}{(2n+1)!}a_0$$ for $n \ge 1 $
I know it says that we can do this by multiply both the top and bottom of the RHS by $2^nn!$
But I am still having some issues seeing how exactly this accomplishes that? Any help? Thank you
$$\begin{align}\frac{(-1)^n}{(3\times 5\times\cdots\times (2n+1))\cdot n!}&=\frac{(-1)^n}{(3\times 5\times\cdots\times (2n+1))\cdot n!}\cdot\frac{\color{red}{2\times 4\times \cdots\times (2n)}}{2\times 4\times \cdots\times (2n)}\\&=\frac{(-1)^n\cdot \color{red}{2^n\cdot n!}}{(1\times 2\times 3\times\cdots\times (2n+1))\cdot n!}\\&=\frac{(-1)^n\cdot 2^n}{(2n+1)!}\end{align}$$
Here, note that $$2\times 4\times \cdots \times(2n)=(2\cdot 1)\times (2\cdot 2)\times\cdots\times (2\cdot n)=2^n\cdot n!.$$