Help understanding the solution to #83 Section 9.1 Calculus 9e, by Larson. 
Find $a_n$ of  $1, \; \frac{-1}{1\times 3}, \; \frac{1}{1\times3\times5}, \frac{-1}{1\times 3\times 5 \times 7}$

I solved it as $\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!}$
The solution guide says: $\dfrac{\left(-1\right)^{n-1}2^n n!}{\left(2n\right)!} $
I am unable to figure out how $2^n$ gets into the numerator. 
if you multiply $\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!}$ by $\dfrac{2n}{2n}$ results in $\dfrac{\left(-1\right)^{n-1}2n}{\left(2n\right)!}$
Please, help me understand how $2n$ goes to $2^n n!$
Have learned from the comments that the denomator is $\left(2n-1\right)!!$  Sorry, still not sure how to solve this problem.  This is my first time learning calculus and haven't seen any double factorials.  Hate this about this Larson Calculus book, great problems, but they are like Sherlock Novels, always some missing knowledge not showen that is needed to solve the later problems.
 A: In the denominator, we note that there's a product of odd numbers and at the numerator there's an alternate sign $\pm$  then $a_n$ must be on the form $a_n=\frac{(-1)^{n-1}}{1\times 3\times \cdots\times (2n-1)}$.
Now, let's multiply both numerator and denominator by $2\times 4\times \cdots\times (2n)$
$$a_n=\frac{(-1)^{n-1}}{1\times 3\times \cdots\times (2n-1)}=\frac{(-1)^{n-1}2\times 4\times \cdots\times (2n)}{1\times2\times3\times 4\times \cdots\times(2n-1)\times (2n)}\\=\frac{(-1)^{n-1}(2\times 1)\times( 2\times 2)\times\cdots\times(2\times n)}{(2n)!}=\frac{(-1)^{n-1}2^nn!}{(2n)!}.$$
A: It is $(2n-1)!!$, not $(2n-1)!$. Use that $(2n)!!=2^n n!$
If you don't know: $(2n-1)!!=(2n-1)(2n-3)\cdot\ldots \cdot3 \cdot 1$ and $(2n)!!=(2n)(2n-2)\cdot \ldots 4 \cdot 2$
Hint: multiple and divide by $(2n)!!$, note that $(2n-1)!! \cdot (2n)!!= (2n)!$
Full solution:

$\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!!}=\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!!} \cdot \dfrac{(2n)!!}{(2n)!!} = \dfrac{\left(-1\right)^{n-1}(2n)!!}{\left(2n-1\right)!!(2n)!!}=\dfrac{\left(-1\right)^{n-1}2^n n!}{\left(2n-1\right)!!(2n)!!}=\dfrac{\left(-1\right)^{n-1}2^n n!}{(2n)!}$

Proof that $(2n)!!=2^n n!$:

\begin{align*}(2n)!!&=(2n)(2n-2)\cdot \ldots \cdot 4 \cdot 2\\ &= (2n)(2(n-1))\cdot \ldots \cdot (2\cdot 2) \cdot (2\cdot 1)\\
&=2^n (n(n-1)\cdot \ldots \cdot 2 \cdot 1) = 2^n n!\end{align*}

Proof that $(2n-1)!! \cdot (2n)!!=(2n)!$:

$\underbrace{(2n-1)(2n-3)\cdot\ldots \cdot3 \cdot 1}_{(2n-1)!!}\cdot \underbrace{ (2n)(2n-2)\cdot \ldots 4 \cdot 2}_{(2n)!!}= (2n)(2n-1)\cdot \ldots \cdot 4 \cdot 3 \cdot 2 \cdot 1=(2n)!$

A: No different results here but a somewhat different way of working through it.
\begin{array}{|c|c|c|c|c|}
\hline
\mathbf{n}&1& 2 &3& 4 & 5\\
\hline
a_n&1&\frac{-1}{1\cdot3}&\frac{1}{1\cdot3\cdot5}&\frac{-1}{1\cdot3\cdot5\cdot7}&\frac{1}{1\cdot3\cdot5\cdot7\cdot9}\\
\end{array}
Terms like $1 \cdot 3 \cdot 5 \cdot 7$ are similar to factorials, except they are
missing the even numbers. But notice that $2^2 \cdot2!$ = $2\cdot4$, that $2^3 \cdot3! = 2\cdot4\cdot6$, that $2^4\cdot4! = 2\cdot 4\cdot6\cdot8$, and so on. So for example $1\cdot3\cdot5\cdot7\cdot 2^33! = 7!$ and  $1\cdot3\cdot5\cdot7\cdot9\cdot 2^44! = 9!$. Let's multiply each $a_n$ by $$\frac{2^{n-1}(n-1)!}{2^{n-1} (n-1)!}$$ so that each denominator becomes a factorial. Each $a_n$ multiplies out to $$\frac{(-1)^{n-1}2^{n-1}(n-1)!}{(2n-1)!}$$ We know that $$\frac{2^{n-1}(n-1)!}{(2n-1)!}=\frac{2^nn!}{2n!}$$ when $n\ge1$. So$$\frac{(-1)^{n-1}2^{n-1}(n-1)!}{(2n-1)!}=\dfrac{\left(-1\right)^{n-1}2^n n!}{\left(2n\right)!}$$
