Is this equation for $2^k!$ correct? I couldn't find any equation for $2^k!$ so I came up with an equation that appears to work for the factorial of a power of $2$. However, I'm having problems proving it. My equation:
$$
\def\x{\times}
\def\prodf#1{\prod\limits_{n=2}^{#1}\prod\limits_{i=2}^{2^{n-1}}(2i-1)}
\def\ptwof#1{2^{2^#1-1}}
\def\foffset{\prod_{x=2^f+1}^{2^{f+1}}}
2^k! = \ptwof{k}\x\prodf{k}
$$
Induction is the obvious way to go:
$$
\begin{aligned}
k=0:&\\
&2^k!=1!=1=2^{1-1}\x1=\ptwof{k}\x\prodf{k}\qquad\checkmark\\
\text{assume}&\text{true for } k=f\\
k=f+&1:\\
&2^{f+1}!=2^f!\x\foffset=\ptwof{f}\x\prodf{f}\x\foffset\\
&_{\text{unsure if this is true; did I unpack }\prodf{f}\text{ right?}}\\
&=\ptwof{f}\x(2^f+1)\x(2^f+2)\x\cdots\x(2^{f+1})\x3^{f-1}\x5^{f-2}\x7^{f-2}\x\cdots\\
&=\quad?
\end{aligned}
$$

I got here because
$$(2j)! = j!\x2^k\x3\x5\x7\x\cdots\x(2j-3)\x(2j-1)\\$$
So if $n=2^k$, then we can keep using that equation to replace $j!$
After trying to do, I got
$$
\ptwof{k}\x3^{p_1}\x5^{p_2}\x\cdots
$$
by testing it out for some numbers and re-arranging stuff (ie I haven't proven this correct, but I guessed it).
The powers of the odd numbers were: 
$$
\begin{array}{|c|c}\hline
    \text{power} &p &p-1 &p-1 &p-2 &p-2 &p-2 &p-2 &p-3 \\ \hline
    \text{number}&3 &5   &7   &9   &11  &13  &15  &17 \\ \hline
\end{array}\cdots
$$
Where $p=k-1$ (for $k>2$; we stop when $p-j$ is $0$)
Basically, before we move $2^i$ numbers right, the $2^i$ next numbers each get the power $p-i$, where $i$ is the number of times we moved right.
For at least up to $2^{5}!$, $\displaystyle\prodf{k}$ is the formula for the multiplying by the odd numbers.
 A: Your formula is correct. The key point is to get the correct order for the Sylow $2$-subgroup of $S_{2^{k}},$ and you have done this. After that, the other term in your product is clearly the odd part of the order of $S_{2^{k}}.$ The proof that the order of  Sylow $2$-subgroup of $S_{2^{k}}$ is what you claim can be done by induction.  You can show that $S_{2^{k-1}} \wr S_{2}$ has odd index in $S_{2^{k}},$ and then everything follows.
Later edit in response to comment: Here's how to get the power of $2$ right without using groups (its more or less what was done in the question): for $1 \leq j \leq 2^{k-2}-1,$ the power of $2$ dividing $2j + 2^{k-1}$ is the same as the power of $2$ dividing $2j.$ 
But for $j = 2^{k-2}$ the power of $2$ dividing $2j +2^{k-1}$ is $k,$ which is one greater than the power of $2$ dividing $2^{k-1}.$ Hence the power of $2$ dividing $2^{k}!$ is the same as the power of $2$ dividing $2 \times (2^{k-1}!)^{2}.$ Then by induction this power is the $(2^{k}-1)$ power.
