How is the factorial of $(2k)!=2^k k! 1.3.5 .....(2k-1)$ I want to figure out  how $(2k)!=2^k \cdot k! \cdot 1\cdot 3\cdot 5\cdot \ldots\cdot(2k-1)$ I know that $(2k)!=2k(2k-1)(2k-2)\cdot \ldots \cdot1$. I tried to figure it but I cant. How does one get from the left side of the equation to the right side? Is there an intuitive explanation as for why this makes sense?
 A: Intuitive explanation
$$\begin{array}{rcl}
(2k)! &=& 1 \cdot 2 \cdot 3 \cdot 4 \cdots (2k) \\
&=& (2 \cdot 4 \cdot 6 \cdots (2k)) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1)) \\
&=& 2^k (1 \cdot 2 \cdot 3 \cdots k) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1)) \\
&=& 2^k (k!) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1)) \\
\end{array}$$

Proof
Use Mathematical induction on $k$.
Let $P(n)$ be the proposition that $(2n)! = 2^n (n!) \cdot (1 \cdot 3 \cdot 5 \cdots (2n-1))$, where $n$ is a natural number.
When $n=0$, $\text{LHS}= 0! = 1$ and $\text{RHS}=1$.
Assume $P(k)$ be true, where $k$ is a natural number.
Then, for $P(k+1)$:
$$\begin{array}{rcl}
\text{LHS} &=& (2(k+1))! \\
&=& (2k)!(2k+1)(2k+2) \\
&=& 2^k (k!) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1))(2k+1)(2k+2) \\
&=& 2^{k+1} (k!) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1))(2k+1)(k+1) \\
&=& 2^{k+1} (k!) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1) \cdot (2k+1))(k+1) \\
&=& 2^{k+1} (k!(k+1)) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1) \cdot (2k+1)) \\
&=& 2^{k+1} (k+1)! \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1) \cdot (2k+1)) \\
&=& \text{RHS}
\end{array}$$
Thus, by the principle of Mathematical Induction, $P(n)$ is true for all natural number $n$.
A: Factor out all even numbers from $(2k)!\,$ You obtain
$$\bigl(2\cdot 4\cdot6\cdots(2k)\bigr)\bigl(1\cdot3\cdot5\cdots(2k-1)\bigr)=2^k(1\cdot2\cdot3\cdots k)\bigl(1\cdot3\cdot5\cdots(2k-1)\bigr)$$
